Number System - Study Mode

[#466] If n = 1 + x, where x is the product of four consecutive positive integers, then which of the following is/are true ? I. n is odd II. n is prime III. n is a perfect square
Correct Answer

(C) I and III only

Explanation

Solution: Let n = 1 + x = 1 + m (m + 1) (m + 2) (m + 3), where m is a positive integer. Then, clearly two of m, (m + 1), (m + 2), (m + 3) are even and so their product is even. Thus, x is even and hence n = 1 + x is odd Also, n = 1 + m (m + 3) (m + 1) (m + 2)
= 1 + (m 2 + 3m) (m 2 + 3m + 2) ⇒ n = 1 + y (y + 2), where m 2 + 3m = y ⇒ n = 1 + y 2 +2y = (1 + y) 2 , which is a perfect square. Hence, I and III are true.

[#467] If the seven digit number 876p37q is divided by 225, then the value of p and q respectively are :
Correct Answer

(C) 0 and 5

Explanation

Solution: 225 = 9 × 25, where 9 and 25 are co-primes. So, a number is divisible by 225 if it is divisible by both 9 and 25 Given number is divisible by 25, only if 7q is divisible by 25, i,e, if q = 5 Sum of digits of given number = (8 + 7 + 6 + p + 3 + 7 + 5) = 36 + p, which must be divisible by 9 This is possible if p = 0 Hence, p = 0, xa0 q = 5

[#468] What is the remainder when 2 31 is divided by 5 ?
Correct Answer

(C) 3

Explanation

Solution: 2 31 = 2 × 2 30 = 2 × (2 2 ) 15 = 2 × 4 15 When n is odd, (x n + a n ) is divisible by (x + a) ∴ (4 15 + 1 15 ) is divisible by (4 + 1) ⇒ (4 15 + 1) is divisible by 5 ⇒ (2 30 + 1) is divisible by 5 ⇒ On dividing 2 30 by 5, we get (5 - 1) i.e., 4 as remainder. ∴ Remainder obtained on dividing 2 31 by 5 = Remainder obtained in dividing (2 × 4) i.e., 8 by 5 = 3

[#469] (800 ÷ 64) × (1296 ÷ 36) = ?
Correct Answer

420

Explanation

Solution: $$eqalign{
& left( {800 div 64}
ight) imes left( {1296 div 36}
ight) cr
& = frac{{800}}{{64}} imes frac{{1296}}{{36}} cr
& = 450 cr} $$

[#470] The total number of 3 digit numbers which have two or more consecutive digits identical is:
Correct Answer

(A) 171

Explanation

Solution: In each set of 100 numbers, there are 10 numbers whose tens digit and unit digit are same. Again in the same set there are 10 numbers whose hundreds and tens digits are same. But one number in each set of 100 numbers whose Hundreds, Tens and Unit digit are same as 111, 222, 333, 444 etc
Hence, there are exactly (10 + 10 - 1) = 19 numbers in each set of 100 numbers. Further there are 9 such sets of numbers Therefore such total numbers = 19 × 9 = 171 Alternatively, 9 × 10 × 10 - 9 × 9 × 9 = 900 - 729 = 171