Tips on cracking Aptitude Questions on Number systems

1. 5 TIPS on cracking Aptitude Questions on Number Systems

2. Tip #1: Factorize the expression • (a + b)(a - b) = (a2 - b2) • (a + b)2 = (a2 + b2 + 2ab) • (a - b)2 = (a2 + b2 - 2ab) • (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) • (a3 + b3) = (a + b)(a2 - ab + b2) • (a3 - b3) = (a - b)(a2 + ab + b2) • (a3 + b3 + c3 - 3abc) = (a + b + c)(a2 + b2 + c2 - ab - bc – ac) • Question: 297 x 297 = ? • Solution: • 297 x 297 = (300 – 3)(300 – 3) = (300 – 3)2 = 3002 + 32 – 2x3x300 • 297 x 297 = 90000 + 9 – 1800 = 88209 • Question: 753 x 753 + 247 x 247 - 753 x 247 = ? • 753 x 753 x 753 + 247 x 247 x 247 • Solution: • Let 753 = a and 247 = b. Then the fraction can be written as: • a2 + b2 – ab = a2 – ab + b2 = 1 = 1 . • a3 + b3 (a + b)(a2 – ab + b2) a + b 1000

3. Tip #2: Use divisibility tests where applicable • Question: How many 3-digit numbers are completely divisible 6 ? • Solution: • Smallest 3-digit number divisible by 6: • 100 is not divisible by 3. • 101 is not divisible by 2. • 102 is divisible by both 2 and 3 and hence, by 6. • Largest 3-digit number divisible by 6: • 999 is not divisible by 2. • 998 is not divisible by 3. • 997 is not divisible by 2. • 996 is divisible by both 2 and 3 and hence, divisible by 6. • Thus, the numbers form an AP with a = 102, d = 6, l = 996, n = ? • 102 + (n – 1)6 = 996 • 6n = 996 – 102 + 6 = 900 • n = 900 / 6 = 150 • Question: If “481d673” is divisible by 9, what should be the smallest whole number in place of d ? • Solution: • For the number to be divisible by 9, the sum of the digits should be divisible by 9. • 4+8+1+6+7+3 = 29 • The whole number just largest than 29 that is divisible by 9 is 36. • Thus, d = 36 – 29 = 7

4. Tip #3: Apply the rules of Binomial Theorem • n • (x + y)n = ΣnCrx(n – r)yr • r=0 • (1 – x)-1= 1 + x + x2 + x3 + …. where x is between 0 and 1 • (1 + x)-1 = 1 – x + x2 – x3 + … where x is between 0 and 1 • xn + 1 is divisible by x + 1 for all odd values of n. • Proof:Let us prove this formula using the method of induction. • When n=1, xn + 1 = x+1 which is divisible by (x+1). When n=2, x2 + 1 is not divisible by (x+1). • Let x2k+1 + 1 be divisible by (x+1). • When n = 2k+3 (the next odd number after 2k + 1) • x2k+3 + 1 = x2(x2k+1 + 1) + (1 - x2) = x2(x2k+1 + 1)+(1+x)(1-x), which is divisible by (x+1). • Thus, xn + 1 is divisible by x + 1 when n is odd. • xn – 1 is divisible by x + 1 only when n is even. [Proof is similar to the one above] • Question: What will be remainder when (6767 + 67) is divided by 68 ? • Solution: • From Binomial Theorem, we know, xn + 1 is divisible by x + 1 only when n is odd. • 6767 + 67 = (6767 + 1) + 66 • Thus, remainder = 66 • Question: Find the value of (0.999)3 correct to 3 decimal places. • Solution: • (0.999) 3 = (1 – 0.001) 3 • = 13 - C(3,1)12(0.001) + C(3,2)1(0.001)2 - (0.001)3 • = 1 - 0.003 = 0.997 (neglecting 3rd and 4th terms since they are « 0.001)

5. Tip #4: Remember the rules for division • Dividend = (Divisor x Quotient) + Remainder • Any recurring decimal can be written as the sum of a non-recurring decimal and a fraction of the recurring portion of the decimal out of (10n– 1). • Ex: 0.125125125…= 0.125 = 125/999 and 7.2341341341… = 7.2 + 341/9999 • Question: Express 0.232323..... As a rational number. • Solution: • 0.232323… = 23/99. • Question: A student mistook the divisor as 12 instead of 21 and obtained 35 as quotient and reminder as 0. What is the correct quotient ? • Solution: • Let the dividend be x. • Then, x / 12 = 35 • x = 12 x 35 • Correct quotient = (12 x 35 / 21) = 4 x 5 = 20. • Question: Find the smallest 6 digit number exactly divisible by 111. • Solution: • Smallest 6 digit number = 100000 • When 100000 is divided by 111, the quotient is 900 and the remainder is 100. • The number would be divisible by 111 if the difference between 99900 (the number just lesser than 100000 that is divisible by 111) and the next divisible number is 111. • Thus, the smallest 6 digit number divisible by 111 is 99900 + 111 = 100011.

6. Tip #5: Memorize the formulae for sum of powers and apply where suitable • 1 + 2 + 3 + ….. + n = n (n + 1) / 2 • 12 + 22 + 32 + ….. + n2 = n (n + 1)(2n + 1) / 6 • 13 + 23 + 33 + …... + n3 = [ n (n + 1) / 2 ] 2 • Question:  (51 + 52 + 53 + ... + 100) = ? • Solution: • (51 + 52 + 53 + ….. + 100) = (1 + 2 + 3 + ….. + 100) – (1 + 2 + 3 + ….. + 50) • = 100(100 + 1) - 50(50 + 1) • 2 2 • = 5050 – 1275 • = 3775. • Question: (22 + 42 + 62 + ... + 202) = ? • Solution: (22 + 42 + 62 + ... + 202) = 22(12 + 22 + 32 + ….. + 102) • = 22 x 10(10 + 1)(2•10 + 1) • 6 • = 4 x 10 x 11 x 21 / 6 • = 1540.

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