### TCS, IBM, Wipro, Infosys Number System Aptitude Question Paper and Answer

The last digit of 1789 is equal to the last digit of 789.The last digits of the powers of 7 follow a cyclic pattern consistingof 4 steps in each cycle. The pattern is 7, 9, 3, 1, 7, 9, 3, 1, .... and so on.The power 89 when divided by 4, the remainder is 1. Hence thelast digit of 1789 is 7. Hence 

The product of two odd numbers is always odd and not even.Hence 

Let the two numbers be 29 A and 29 BHCF × LCM = Product of numbers. ?29 × 4147 = 29A × 29B    ?11 × 13 = A × B ? A = 11; B = 13

1771, when divided by 12, the remainder is 7.1773 when divided by 12, the remainder is 9.1775 when divided by 12, the remainder is 11.1777 when divided by 12, the remainder is 13.?The remainder when the sum of 1771, 1773, 1775 and 1777divided by 12 is = (7 + 9 + 11 + 13) = 4040 when divided by 12 the remainder is 4?The remainder is 4. Hence 

The Sum of even three terms = 3 + 2 – 5 = 0.?Sum of 48 terms = 0Hence sum of 49th and 50th terms = 3 + 2 = 5. Hence 

We know 63 = 216? 63 when divided by 216, the remainder is 1? (63)4 when divided by 216, the remainder is 112, i.e. 1.? 6 12 × 6 when divided by 216, the remainder is (1 × 6), i.e. 6.Hence 

Let n = 2, ?2 (3) (4) = 24, which is divisible by 24Let n = 3 ? 3 (9 – 1) = 3 (8) = 24, which is divisible by 24.Let n = 2, 54 – 1 = 625 – 1 = 624, which is divisible by 13.Let n = 1, 52 + 1= 26 which is divisible by 13Hence 

The number of divisors of a given number N (including 1 and thenumber it self) isN = am bn cp,Where a, b, c are prime are (1 + m) (1 + n) (1 + p)

x n – yn is divisible by (x-y) when n is odd is true.x n – yn is divisible by x+ y when n is even and not odd.x n – yn is divisible by (x+y) when n is odd and not even.Hence 

The last digit of 2563 is 5.The last digit of 6325 is 3.The last digit of the product is 5. Hence 

The numbers less than a lakh = 45Eliminating the case where all digits are 0, the answer is 45– 1.Hence 

Let 701 ? x (mod y) and 665 ? x (mod y)Subtracting we get, 36 ? 0 (mod y)i.e 36 is divisible by ‘y’. Hence 

Given that a > 0 and b < 0Adding, we get a – b > 0. Hence 

f(x) = 3 + 8x – x2f?(x) = 8 – 2xf??(x) = –2 < 0 ? it is a maximum.For a maximum f? (x) = 0? 8 – 2x = 0 ? x = 8/2 = 4? It is maximum at x = 4The maximum value = 3 + 8 (4) – (4)2 =19Hence