Divisibility rules:
-> A number is divisible by 2 if and only if the last digit is divisible by 2.
-> A number is divisible by 3 if and only if the sum of the digits is divisible by 3.
-> A number is divisible by 4 if and only if the last 2 digits is a number divisible by 4.
-> A number is divisible by 5 if and only if the last digit is divisible by 5.
-> A number is divisible by 6 if and only if it is divisible by 2 and 3.
-> A number is divisible by 8 if and only if the last 3 digits is a number divisible by 8.
-> A number is divisible by 9 if and only if the sum of the digits is divisible by 9.
-> A number is divisible by 10n if and only if the number ends in n zeros.
-> A number is divisible by 11 iff the sum of every other digit minus the sum of the rest of the digits is divisible by 11.
-> To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.
-> If n is even , n(n+1)(n+2) is divisible by 24
Number Tips
-> For three positive numbers a, b ,c
(a+b+c) * (1/a+1/b+1/c)>=9
-> For any positive integer n
2<= (1+1/n)^n <=3
-> The greatest common divisor is found by looking at the prime factorizations or using the Euclidean algorithm.
-> The least common multiple of a and b is found by looking at the prime factorizations or (ab)/gcd(a,b).
-> Two numbers are said to be relatively prime in the greatest common factor is 1.
-> If gcd(a, b)=d, then there exist integers x and y so that ax+by=d.
-> If d divides both a and b, then d divides a+b and d divides a-b.a=b mod m iff m divides a-b iff a and b both have the same remainder when divided by m.
-> ap-1 = 1 mod p (a is not a multiple of p)af(m)=1 mod m ( gcd(a, m) =1)
-> If a probability experiment is repeated n times and the probability of success in one trial is p, then the probability of exactly r successes in the n trials is nCr (p)r(1-p)(n-r).
-> The number of zeros at the end of n! is determined by the number of 5’s. To find this you do the following process: n/5 = n1 and some remainder. Drop the remainder and compute n1/5 = n2 plus some remainder. Drop the remainder and compute n2/5 = n3 plus some remainder, etc. The number of zeros is n1+n2+n3+n4...
-> The sum of any consecutive integers k through n, with n being the larger,simply use this equation:(n+k)(n-k+1)------------ 2
-> a^2+b^2+c^2 >= ab+bc+ca If a=b=c , then the equality holds in the above. -> a^4+b^4+c^4+d^4 >=4abcd
-> (n!)^2 > n^n (! for factorial)
-> If a+b+c+d=constant , then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s .
-> Consider the two equations a1x+b1y=c1 a2x+b2y=c2
-> (m+n)! is divisible by m! * n! .Then , If a1/a2 = b1/b2 = c1/c2 , then we have infinite solutions for these equations. If a1/a2 = b1/b2 <> c1/c2 , then we have no solution for these equations.(<> means not equal to ) If a1/a2 <> b1/b2 , then we have a unique solutions for these equations..
-> When a three digit number is reversed and the difference of these two numbers is taken , the middle number is always 9 and the sum of the other two numbers is always 9 .
-> Let 'x' be certain base in which the representation of a number is 'abcd' , then the decimal value of this number is a*x^3 + b*x^2 + c*x + d
Combinatorics Tips
-> The number of arrangements of n objects is n!-> The number of arrangements of r out of n objects is nPr = n!/(n-r)!
-> The number of arrangements of n objects in a circle is (n-1)!
-> The number of arrangements of n objects on a key ring is (n-1)!/2
-> The number of arrangements of n objects with r1 of type 1, r2 of type 2, ..., ri of type i is n!/(r1!r2!...ri!)
-> The number of ways of choosing n out of r objects is nCr = n!/((n-r)! r!)
-> The number of distributions of n distinct objects in k distinct boxes is kn.
-> The number of ways of distributing n identical objects in k distinct boxes is (n+k-1)Cn.
-> The sum of the coefficients of the binomial expression (x+y)n is 2n.
-> To find the sum of the coefficients of a power of any polynomial, replace the variables by 1.
-> When solving an equation for integer solutions, it is important to look for factoring. Important factoring forms:
1. .a2-b2=(a-b)(a+b)
2. a3-b3=(a-b)(a2+ab+b2)
3. an-bn=(a-b)(an-1+an-2b+an-3b2+...+abn-2+bn-1)
4. a3+b3=(a+b)(a2-ab+b2)
5. If n is odd, an+bn=(a+b)(an-1-an-2b+an-3b2-...-abn-2+bn-1) (alternate signs)
More Tips on Geometry
-> If any parallelogram can be inscribed in a circle , it must be a rectangle.
-> If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i:e oblique sides equal).
-> For an isosceles trapezium , sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order) .
-> Area of a regular hexagon : root(3)*3/2*(side)*(side)
-> For any quadrilateral whose diagonals intersect at right angles , the area of the quadrilateral is 0.5*d1*d2, where d1,d2 are the lenghts of the diagonals.
-> In any triangle a/SinA = b/SinB =c/SinC=2R , where R is the circumradius
-> In an isosceles triangle , the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.
-> In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.
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Cyclic Quadrilateral
-> For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2
-> For a cyclic quadrilateral , the measure of an external angle is equal to the measure of the internal opposite angle.
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Quadrilateral
-> If a quadrilateral circumscribes a circle , the sum of a pair of opposite sides is equal to the sum of the other pair . -> The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.
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Area of a triangle
1/2*base*altitude = 1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = root(s*(s-a)*(s-b)*(s-c)) where s=a+b+c/2 =a*b*c/(4*R)
where R is the CIRCUMRADIUS of the triangle = r*s ,
where r is the inradius of the triangle .
In any triangle a=b*CosC + c*CosB b=c*CosA + a*CosC c=a*CosB + b*CosA
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Area of a hexagon
Area of a hexagon = root(3) * 3 * (side)^2
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Area of a trapezium
Area of a trapezium = 1/2 * (sum of parallel sids) * height = median * height
where median is the line joining the midpoints of the oblique sides.
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Co-ordinates of Centroidi
The coordinates of the centroid of a triangle with vertices (a,b) (c,d) (e,f) is((a+c+e)/3 , (b+d+f)/3) .
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Ratio of Radii of Corcumcircle & In-circle
The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1 .
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Area of Parallelogram
Area of a parallelogram = base * height
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APPOLLONIUS THEOREM:
In a triangle , if AD be the median to the side BC , then AB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2) .
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Similar Cones
For similar cones , ratio of radii = ratio of their bases. The HCF and LCM of two nos. are equal when they are equal .
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Pyramids
Volume of a pyramid = 1/3 * base area * height
Geometry Tips
-> The area of a square is one-half the the square of the length of a diagonal
-> A line parallel to a side of a triangle that intersects the other two sides divides them proportionally.
-> The angle bisector of an angle divides the opposite side proportional to the other two sides.
-> The angle bisectors intersect at a point called the incenter, that is the center of the inscribed circle.
-> The length of the radius of the inscribed circle is equal to the area of the triangle divided by the semiperimeter.
-> The perpendicular bisectors meet at a point, called the circumcenter, that is the center of the circumscribed circle.
-> The altitudes meet at a point called the orthocenter.
-> The medians intersect at a point (called the centroid) that is two-thirds of the distance from each vertex to the midpoint of the opposite side.
-> The coordinates of the centroid can be found by averaging the coordinates of the vertices.
-> (Ceva’s Theorem) In triangle ABC, with J on AB, K on BC, and L on CA, the segments AK, BL and CJ are concurrent (intersect in one point) if and only if (AJ/JB)(BK/KC)(CL/LA) = 1.-
-> (Ceva’s Theorem Trig Form) In triangle ABC, with J on AB, K on BC, and L on CA, the segments AK, BL and CJ are concurrent (intersect in one point) if and only if (sin(BAK)/sin(KAC))(sin(CBL)/sin(LBA))(sin(ACJ)/sin(JCB)) = 1.
-> The height of an equilateral triangle is sÃ3/2 (So the area is s2 Ã3/4)
-> A regular hexagon inscribed in a circle can be divided into six congruent equilateral triangles, where the length of a side is equal to the radius of the circle. Therefore the area of a regular hexagon inscribed in a circle of radius r is: A=r23Ã3/2.
-> The midpoint of the hypotenuse of a right triangle is equidistant to all three vertices.
-> When an altitude is drawn to the hypotenuse of a right triangle, three similar triangles are formed, and the following hold:
a. The square of the length of the altitude is equal to the product of the lengths of the segments of the hypotenuse.
b. When the length of each leg is squared it equals the product of the hypotenuse and the segment on the hypotenuse adjacent to that leg.
-> The sum of the exterior angles of a polygon, one at each vertex, is 360.
-> The interior angles of a regular polygon with n sides are 180-(360/n) each.
-> Two tangent segments drawn to a the same circle from the same point are equal.
-> The figure made by connecting a common external tangent, two radii, and a segment
connecting the two centers is a trapezoid with two right angles. (Often you divide into a rectangle and a right triangle to solve for missing lengths.)
-> Two chords have the same lengths if and only if they cut off equal length arcs.
-> Two chords are have the same lengths if and only if they are the same distance from the center.
-> A diameter drawn through a chord is perpendicular to the chord if and only if it bisects the chord and the arc.
-> The measure of an angle with its vertex at the center of a circle is equal to the measure of the intercepted arc.
-> The measure of an inscribed angle (that is an angle with its vertex on the circle) is one-half the intercepted arc.
-> If two inscribed angles intercept the same arc, then the angles have the same measure.
-> The measure of the angle made by a chord and a tangent is one-half the measure of the intercepted arc.
-> An angle inscribed in a semicircle is a right angle.
-> A triangle inscribed in a circle is a right triangle if and only if one of the sides is a diameter.
-> The measure of the angle formed by two intersecting chords is equal to one-half the sum of
the measures of the intercepted arcs of the angle and its vertical angle.
-> The measure of the angle formed by two secants, two tangents, or a secant and a tangent drawn from a point outside a circle is equal to one-half the difference of the measures of the intercepted arcs.
-> If two chords intersect, the product of the lengths of the two parts of one chord is equal to the product of the lengths of the parts of the other chord.
-> When two secant segments are drawn to a circle from an external point, the product of the lengths of one secant segment and its external segment equals the product of the lengths of the other secant segment and its external segment.
-> When a secant segment and a tangent segment are drawn to a circle from an external point, the product of the lengths of the secant segment and its external segment is equal to the square of the length of the tangent segment.
-> In a cyclic quadrilaterals (If a quadrilateral is inscribed in a circle, it is called a cyclic quadrilateral),opposite angles are supplementary.(Hence, rectangles or squares are the only parallelograms that can be inscribed in a circle.)
-> (Ptolemey’s Theoren) The product of the lengths of the diagonals is equal to the sum of the products of the lengths of the two pairs of opposite sides.
-> When a circle is circumscribed about an equilateral triangle, and chords are drawn from any point on the circle to the three vertices of the triangle, then the length of the longest chord is equal to the sum of the lengths of the other two chords.
-> Euler’s formula for polyhedron: f+v-e=2 (The number of faces plus the number of vertices minus the number of edges equals 2.
-> There are only 5 regular polyhedron: tetrahedron (4 triangles), hexahedron (a cube, 6 squares), Octahedron (8 triangles), dodecahedron (12 pentagons), and icosahedron (20 triangles).
-> If the scale factor of two similar polygons is a:b, the ratio of their areas is a2:b2.
-> If the scale factor of two similar solids is a:b, the ratio of the lengths of two corresponding edges is a:b.
-> If the scale factor of two similar solids is a:b. the ratio of their surface areas is a2:b2.
-> If the scale factor of two similar solids is a:b. the ratio of their volumes is a3:b3.
-> Heron’s Formula: If a triangle has sides of length a, b and c and s=(a+b+c)/2, then the area of the triangle is sqrt(s*(s-a)*(s-b)*(s-c)).
-> If a rectangle is of a size that when a square is cut off of an end the remaining rectangle is similar to the original, the rectangle is said to be a “golden rectangle” and its sides are said to be in the “golden ratio”. This ratio is 1:x, where x is the positive root of x2-x-1=0. (x=(1+sqrt(5))/2)(This is also the limit as k goes to infinity of two consecutive terms of the Fibonacci sequence)
-> The area of the triangle that has vertices with coordinates (x1, y1), (x2, y2), and (x3, y3) can be found by the absolute value of the determinant of the matrix:x1 y1 1x2 y2 1x3 y3 1This can be generalized to any polygon.
More Tips on Geometry
Quantitative Ability-3
41. The relationship between base 10 and base ‘e’ in log is given by log10N = 0.434 logeN
42. WINE and WATER formula
Let Q - volume of a vessel, q - qty of a mixture of water and wine be removed each time from a mixture, n - number of times this operation is done and A - final qty of wine in the mixture, then,
A/Q = (1-q / Q)^n
43. Pascal’s Triangle for computing Compound Interest (CI)
The traditional formula for computing CI is
CI = P*(1+R/100)^N – P
Using Pascal’s Triangle,
Number of Years (N)
-------------------
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
… 1 .... .... ... ... ..1
Eg: P = 1000, R=10 %, and N=3 years. What is CI & Amount?
Step 1:
Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331
The coefficients - 1,3,3,1 are lifted from the Pascal's triangle above.
Step 2:
CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331 (leaving out first term in step 1)
If N =2, we would have had,
Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs.1210
CI = 2 * 100 + 1* 10 = Rs.210
44. Suppose the price of a product is first increased by X% and then decreased by Y% , then, the final change % in the price is given by:
Final Difference% = X - Y - XY/100
Eg) The price of a T.V set is increased by 40 % of the cost price and then is decreased by 25% of the new price. On selling, the profit made by the dealer was Rs.1000. At what price was the T.V sold?
Applying the formula,
Final difference% = 40 – 25 - (40*25/100) = 5 %.
So if 5 % = 1,000
Then, 100 % = 20,000.
Hence, C.P = 20,000
& S.P = 20,000+ 1000= 21,000
45. Where the cost price of 2 articles is same and the mark up % is same, then, marked price and NOT cost price should be assumed as 100.
46.
-> Where ‘P’ represents principal and ‘R’ represents the rate of interest, then, the difference between 2 years’ simple interest and compound interest is given by P * (R/100)2
-> The difference between 3 years’ simple interest and compound interest is given by (P * R2 *(300+R))/1003
47.
-> If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.
-> If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.
-> If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time
-> If A can finish a work in X time and B in Y time and A, B & C together in S time then
· C can finish that work alone in (XYS)/ (XY-SX-SY)
· B+C can finish in (SX)/(X-S); and
· A+C can finish in (SY)/(Y-S)
48. In case ‘n’ faced die is thrown k times, then, probability of getting atleast one more than the previous throw = nC5/n5
49.
-> When an unbiased coin is tossed odd no. (n) of times, then, the no. of heads can never be equal to the no. of tails i.e. P (no. of heads=no. of tails) = 0
-> When an unbiased coin is tossed even no. (2n) of times, then, P (no. of heads=no. of tails) = 1-(2nCn/22n)
50. Where there are ‘n’ items and ‘m’ out of such items should follow a pattern, then, the probability is given by 1/m!
Eg) 1. Suppose there are 10 girls dancing one after the other. What is the probability of A dancing before B dancing before C?
Here n=10, m=3 (i.e. A, B, C)
Hence, P (A>B>C) = 1/3!
= 1/6
Eg)2. Consider the word ‘METHODS’. What is the probability that the letter ‘M’ comes before ‘S’ when all the letters of the given word are used for forming words, with or without meaning?
P (M>S) = 1/2!
= 1/2
51. CALENDAR
-> Calendar repeats after every 400 years.
-> Leap year- it is always divisible by 4, but century years are not leap years unless they are divisible by 400.
-> Century has 5 odd days and leap century has 6 odd days.
-> In a normal year 1st January and 2nd July and 1st October fall on the same day. In a leap year 1st January 1st July and 30th September fall on the same day.
-> January 1, 1901 was a Tuesday.
52.
-> For any regular polygon, the sum of the exterior angles is equal to 360 degrees, hence measure of any external angle is equal to 360/n (where n is the number of sides)
-> For any regular polygon, the sum of interior angles =(n-2)*180 degrees
So measure of one angle is (n-2)/n *180
-> If any parallelogram can be inscribed in a circle, it must be a rectangle.
-> If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i.e. oblique sides equal).
53. For an isosceles trapezium, sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides (i.e. AB+CD = AD+BC, taken in order)
54.
-> For any quadrilateral whose diagonals intersect at right angles, the area of the quadrilateral is
0.5*d1*d2, where d1, d2 are the length of the diagonals.
-> For a cyclic quadrilateral, area = root((s-a) * (s-b) * (s-c) * (s-d)), where s=(a + b + c + d)/2
Further, for a cyclic quadrilateral, the measure of an external angle is equal to the measure of the interior opposite angle.
-> Area of a Rhombus = Product of Diagonals/2
55. Given the coordinates (a, b); (c, d); (e, f); (g, h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for
[(a + e)/2, (b + f)/2] = [(c + g)/2, (d + h)/2]
56. Area of a triangle
-> 1/2*base*altitude
-> 1/2*a*b*sin C (or) 1/2*b*c*sin A (or) 1/2*c*a*sin B
-> root(s*(s-a)*(s-b)*(s-c)) where s=(a+b+c)/2
-> a*b*c/(4*R) where R is the circumradius of the triangle
-> r*s ,where r is the inradius of the triangle
57. In any triangle
-> a=b*cos C + c*cos B
-> b=c*cos A + a*cos C
-> c=a*cos B + b*cos A
-> a/sin A=b/sin B=c/sin C=2R, where R is the circumradius
-> cos C = (a^2 + b^2 - c^2)/2ab
-> sin 2A = 2 sin A * cos A
-> cos 2A = cos^2 (A) - sin^2 (A)
58. The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1
59. Appollonius Theorem
In a triangle ABC, if AD is the median to side BC, then
AB2 + AC2 = 2(AD2 + BD2) or 2(AD2 + DC2)
60.
-> In an isosceles triangle, the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.
-> In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.
61. The quadrilateral formed by joining the angular bisectors of another quadrilateral is always a rectangle.
62. Let W be any point inside a rectangle ABCD, then,
WD2 + WB2 = WC2 + WA2
63. Let a be the side of an equilateral triangle, then, if three circles are drawn inside this triangle such that they touch each other, then each circle’s radius is given by a/(2*(root(3)+1))
64.
-> Distance between a point (x1, y1) and a line represented by the equation ax + by + c=0 is given by ax1+by1+c/Sq(a2+b2)
-> Distance between 2 points (x1, y1) and (x2, y2) is given by Sq((x1-x2)2+ (y1-y2)2)
65. Where a rectangle is inscribed in an isosceles right angled triangle, then, the length of the rectangle is twice its breadth and the ratio of area of rectangle to area of triangle is 1:2.
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Quantitative Ability-2
21. x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)
22. e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
Note: 2 <>23. log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]
24. (m + n)! is divisible by m! * n!
25. When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.
26. Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)
27. To Find Square of a 3-Digit Number
Let the number be XYZ
Steps
a.Last digit = Last digit of Sq(Z)
b. Second last digit = 2*Y*Z + any carryover from STEP 1
c. Third last digit 2*X*Z+ Sq(Y) + any carryover from STEP 2
d. Fourth last digit is 2*X*Y + any carryover from STEP 3
e. Beginning of result will be Sq(X) + any carryover from Step 4
Eg) Let us find the square of 431
Step
a. Last digit = Last digit of Sq(1) = 1
b. Second last digit = 2*3*1 + any carryover from STEP 1=6+0=6
c. Third last digit 2*4*1+ Sq(3) + any carryover from STEP 2 = 8+9+0 = 17 i.e. 7 with carry over of 1
d. Fourth last digit is 2*4*3 + any carryover from STEP 3 = 24+1 = 25 i.e. 5 with carry over of 2
e. Beginning of result will be Sq(4) + any carryover from Step 4 = 16+2 = 18
THUS SQ(431) = 185761
28. If the answer choices provided are such that the last two digits are different, then, we need to carry out only the first two steps only.
-> The sum of first n natural numbers = n(n+1)/2
-> The sum of squares of first n natural numbers is n(n+1)(2n+1)/6
-> The sum of cubes of first n natural numbers is (n(n+1)/2)2/4
-> The sum of first n even numbers= n (n+1)
-> The sum of first n odd numbers= n2
29. If a number ‘N’ is represented as a^x * b^y * c^z… where {a, b, c, …} are prime numbers, then
-> the total number of factors is (x+1)(y+1)(z+1) ....
-> the total number of relatively prime numbers less than the number is N * (1-1/a) * (1-1/b) * (1-1/c)....
-> the sum of relatively prime numbers less than the number is N/2 * N * (1-1/a) * (1-1/b) * (1-1/c)....
-> the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * ...../(x * y *...)
-> Total no. of prime numbers between 1 and 50 is 15
-> Total no. of prime numbers between 51 and 100 is 10
-> Total no. of prime numbers between 101 and 200 is 21
-> The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6
-> The number of rectangles in n*m board is given by n+1C2 * m+1C2
30. If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.
Certain nos. to be remembered
-> 210 = 45 = 322 = 1024
-> 38 = 94 = 812 = 6561
-> 7 * 11 * 13 = 1001
-> 11 * 13 * 17 = 2431
-> 13 * 17 * 19 = 4199
-> 19 * 21 * 23 = 9177
-> 19 * 23 * 29 = 12673
31.
-> The number of squares in n*m board is given by m*(m+1)*(3n-m+1)/6
-> The number of rectangles in n*m board is given by n+1C2 * m+1C2
32. If ‘r’ is a rational no. lying between 0 and 1, then, r^r can never be rational.
33. Certain nos. to be remembered
-> 210 = 45 = 322 = 1024
-> 38 = 94 = 812 = 6561
-> 7 * 11 * 13 = 1001
-> 11 * 13 * 17 = 2431
-> 13 * 17 * 19 = 4199
-> 19 * 21 * 23 = 9177
-> 19 * 23 * 29 = 12673
34. Where the digits of a no. are added and the resultant figure is 1 or 4 or 7 or 9, then, the no. could be a perfect square.
35. If a no. ‘N’ has got k factors and a^l is one of the factors such that l>=k/2, then, a is the only prime factor for that no.
36. To find out the sum of 3-digit nos. formed with a set of given digits
This is given by (sum of digits) * (no. of digits-1)! * 1111…1 (i.e. based on the no. of digits)
Eg) Find the sum of all 3-digit nos. formed using the digits 2, 3, 5, 7 & 8.
Sum = (2+3+5+7+8) * (5-1)! * 11111 (since 5 digits are there)
= 25 * 24 * 11111
=6666600
37. Consider the equation x^n + y^n = z^n
As per Fermat’s Last Theorem, the above equation will not have any solution whenever n>=3.
38. Further as per Fermat, where ‘p’ is a prime no. and ‘N’ is co-prime to p, then, N^(p-1) – 1 is always divisible by p.
39. 145 is the 3-digit no. expressed as sum of factorials of the individual digits i.e.
145 = 1! + 4! + 5!
40.
->Where a no. is of the form a^n – b^n, then,
· The no. is always divisible by a - b
· Further, the no. is divisible by a + b when n is even and not divisible by a + b when n is odd
-> Where a no. is of the form a^n + b^n, then,
· The no. is usually not divisible by a - b
· However, the no. is divisible by a + b when n is odd and not divisible by a + b when n is even
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Quantitative Ability
- Quantitative Ability – POINTS TO REMEMBER
1. If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.
Eg: x3+3x2+2x+6=0 has no positive roots
2. For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.
3. For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)
4. Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be 2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.
5.
a.For a cubic equation ax3+bx2+cx+d=o
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Product of the roots = -d/a
b.For a bi-quadratic equation ax4+bx3+cx2+dx+e = 0
Sum of the roots = - b/a
Sum of the product of the roots taken three at a time = c/a
Sum of the product of the roots taken two at a time = -d/a
Product of the roots = e/a
6. If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)
7. Consider the two equations
a1x+b1y=c1
a2x+b2y=c2
Then,
a. If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.
b. If a1/a2 = b1/b2 <> c1/c2, then we have no solution.
c. If a1/a2 <> b1/b2, then we have a unique solution.
8. Roots of x2 + x + 1=0 are 1, w, w2 where 1 + w + w2=0 and w3=1
9. a + b = a + b if a*b>=0
else, a + b >= a + b
10. The equation ax2+bx+c=0 will have max. value when a<0>0. The max. or min. value is given by (4ac-b2)/4a and will occur at x = -b/2a
11. a. If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4.
b. If for two numbers x*y=k (a constant), then their SUM is MINIMUM if x=y (=root(k)). The minimum sum is then 2*root (k).
12. Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.
13. For any 2 numbers a, b where a>b
a. a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)
b. (GM)^2 = AM * HM
14. For three positive numbers a, b, c
(a + b + c) * (1/a + 1/b + 1/c)>=9
15. For any positive integer n
2<= (1 + 1/n)^n <=3
16. a2 + b2 + c2 >= ab + bc + ca
If a=b=c, then the case of equality holds good.
17. a4 + b4 + c4 + d4 >= 4abcd (Equality arises when a=b=c=d=1)
18. (n!)2 > nn
19. If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s
20. If n is even, n(n+1)(n+2) is divisible by 24
Fast Arithmetic Tips
Checking the result fast
Divisibility by 5.
A number is divisible by 5 iff it ends with either 5 or 0. So you know right away that 123·345=42434 is wrong. It's a different question how to get the right answer (42435).
Use divisibility criteria.
As above, divisibility criteria are useful in establishing whether a result is wrong. Note that this way you never may be sure whether the result is right. The following criteria is available:
Divisibility by 3
Divisibility by 9
Divisibility by 7
Divisibility by 11
A number is divisible by 4 iff its 2-digit tail is divisible by 4. For example, 12345 is not divisible by 4 since 45 is not. However, 12348 is divisible by 4 since 48 is.
A number is divisible by 8 iff its 3-digit tail is divisible by 8. For example, 12345 is not divisible by 8 since 345 is not. However, 12344 is divisible by 8 since 344 is. Indeed 344=320+24, where both operands are divisible by 8.
A number is divisible by 25 if it ends with one of the four: 00, 25, 50, 75. Therefore 234465 is not divisible by 25.
The is a general way of deriving divisibility criteria for numbers such as 7, 11, 13, and other.
Use rude bounds.
127·345=27455 is clearly wrong because 127>100 while 345>300. Therefore, it must be that 127·345>100·300=30000.
220=524288 is clearly wrong because 220=(210)2=10242>10002=1000000. Actually 524288=219.
A practical application.
Use digital roots.
Digital root of an integer is, by definition, the sum of its digits computed recursively until only one digit remains. Thus to compute the digital root of 33572, we calculate first 3 + 3 + 5 + 7 + 2 = 20. The digitl root of 33572 is then 2 = 2 + 0.
The important fact about digital roots is that the digital root of the product of integers equals the product of their digital roots (downsized to a digital root if necessary.)
348·11 = 3828. The digital roots of 348, 11, and 3828 are respectively 6, 2, and 3. Take the product of 6 and 2: 6·2 = 12 with the digital root equal to 1 + 2 = 3, which is exactly the digital root of 3828.
Consider an integer a
a = an10n + an-110n-1 +...+ a1101 + a0
Criteria of divisibility by 9
a is divisible by 9 if and only if the sum of the digits an + an-1 + ... + a1 + a0 is divisible by 9.
Criteria of divisibility by 11
a is divisible by 11 if and only if the alternating sum of the digits (-1)nan + (-1)n-1an-1 + ... -a1 + a0 is divisible by 11.
In what follows I shall be using representation of a number in various base systems. In the decimal (base 10) system a number a is written as
a = (a)10 = an10n + an-110n-1 + ... + a1101 + a0
Generally speaking, for an integer base b the same number a will be written as
a = (a)b = ckbk + ck-1bk-1 + ... + c1b1 + c0
Notation (a)b is meant to underscore the base in which a is expanded. The most customary is the decimal system. But, say, the French still use remnants of the system in base 20. With the advent of computers the binary (base 2) and hexadecimal (base 16) have been thrust into prominence. The octal system (base 8) is very useful in handling lengthy binary representations.
Conversion between different bases is a good exercise that helps test your understanding of relevant techniques.
Theorem
Let b be an integer. Then an integer a
is divisible by (b-1) if and only if the sum of digits in its expansion (a)b is divisible by (b-1)
is divisible by (b+1) if and only if the alternating sum of digits in its expansion (a)b is divisible by (b+1)
Proof
The proof is based on Modulo Arithmetic. Thus we have b-1=0 (mod (b-1)) which implies b=1 (mod (b-1)). Therefore, for every i>0, bi = 1 (mod(b-1)). Multiplying this by ci and summing up for i=0,1, ..., k we get the first assertion.
Similarly, b+1=0 (mod (b+1)) hence b=-1 (mod (b+1)). Therefore, for i>0, bi = (-1)i (mod(b+1)). Multiplying this by ci and summing up for i=0,1, ..., k we get the second assertion.
Example 1
Let a=164. Is it divisible by 4? On the one hand, 164=125+25+2·5+4=(1124)5. The sum of digits is 1+1+2+4=8 and is divisible by 4. Therefore 164 is divisible by 4.
On the other hand, 164=2·81+0·27+0·9+0·3+2=(20002)3. The alternating sum of digits is 2-0+0-0+2=4 and is divisible by 4. Therefore, again, 164 is divisible by 4.
For divisibility by 7 there are two criteria:
a = (a)6 = ck6k+ck-16k-1+...+c161+c0 is divisible by 7 iff the alternating sum of digits (-1)kck+(-1)k-1ck-1+...+c0 is divisible by 7.
a = (a)8=ck8k+ck-18k-1+...+c181+c0 is divisible by 7 iff the sum of digits ck+ck-1+...+c0 is divisible by 7.
Example 2
512=83. Therefore (512)10=(1000)8. This implies that divided by 7 the remainder will be 1. Then, for example, 511 is divisible by 7.
1296=64. Therefore (1296)10=(10000)6. From here (1296+6)10=(10010)6. Its alternating sum of digits is 1-0+0-1+0=0. As a result, 1302 is divisible by
Problem
5109094x171709440000 = 21!, find x.
Solution
I can immediately think of a couple of question:
What is so special (if, of course, anything at all) about the 8th digit from the left? Can I hide another digit and have a meaningful problem?
What is so special about the number 21? If I compute the factorial of another number, say 10, would I be able to retrieve a hidden digit?
With regard to the first question, all digits look the same to me except for the trailing zeros. Zeros at the end of a number indicate the number of factors of 10 this number has. Let pursue this. Where do these factors of 10 come from? The number being a factorial, it's the product
21! = 1·2·3·4·5·...·19·20·21.
Then there are two factors (10 and 20) that contribute zeros to the result and another two (5 and 15) that combine with other even factors (of which there is a plenty) to add another two 0s. At this point, we began utilizing the fact that the given number is known to have certain factors. However, returning to the question asked, I can't think of a way in which other factors of 21! have affected the 8th digit.
As far as my current understanding of the problem goes, there is nothing special about the 8th digit. I also begin suspecting that the choice of 21 in the formulation of the problem is not very important. Indeed it would not be important if the question was to count the number of trailing 0s.
The important thing is that the number is presented as a product of several factors. This is actually the only piece of information that could be gathered from the original formulation. What do I know about factors? They are divisors of a product. Yes, indeed, I have already used properties of multiples of 5 and 10, right? Do I know of other rules concerning division of a number by other numbers? Actually, in school, we do not study a lot about numbers and their features. The study of Arithmetic concentrates on addition and other operations, on the table of multiplication - nothing very generic. And next we plunged into Algebra where numbers have hardly been mentioned at all. Is it also your recollection? (An aside: this is why I disagree with the dictionary definitions of Mathematics as a study of numbers. There is nothing in our education that remotely suggests any study of numbers as such.)
I am lucky, however, to remember one property of numbers divisible by 3 and 9.
A number is divisible by 3 (or 9) iff the sum of its digits is divisible by 3 (or 9).
I start feeling excited. The statement above does not say anything specific about the number of digits or their position. This jibes well with the hunch that selection of the 8th digit was arbitrary. But to apply this statement we have to establish whether the number is divisible by 3 (or 9). It's easy to check that all factorials starting with 3! are divisible by 3 whereas, starting with 6!, all factorials are divisible by 9. Therefore, 21! is divisible by 9. Can we use this?
The sum of digits of 21!, as it's presented in the problem, is (61+x). This is divisible by 9 iff x=2. This must be the missing digit! Division by 3 could also be used for the 8th digit. However, it would fail for the first one.
Summing up:
Given n!, n>5, with one digit removed.Then it is possible to recover the digit.
Please do not jump to conclusions. Remember the problem we just had trying to recover the first digit? May there arise other complications? Sure. There is no way (or, rather, we have not established a way) to tell 0 from 9.
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