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Number System Questions and Answers

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101.

How many of the following numbers are divisible by 132 ?
264, 396, 462, 792, 968, 2178, 5184, 6336

Answer: A

\(132 = 4 \times 3 \times 11\)
So, if the number divisible by all the three number 4, 3 and 11, then the number is divisible by 132 also.
\(264 \Rightarrow 11,3,4\)
\(396 \Rightarrow 11,3,4\)
\(462 \Rightarrow 11,3\)
\(792 \Rightarrow 11,3,4\)
\(968 \Rightarrow 11,4\)
\(2178 \Rightarrow 11,3\)
\(5184 \Rightarrow 3,4\)
\(6336 \Rightarrow 11,3,4\)
Therefore the following numbers are divisible by 132 : 264, 396, 792 and 6336.
Required number of number = 4.

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102.

The sum of three consecutive even numbers is 42. Find the middle number of the three?

Answer: A

Three consecutive even numbers \((2P - 2), 2P, (2P + 2)\)
\((2P - 2) + 2P + (2P + 2) = 42\)
\(6P = 42 => P = 7\)
The middle number is: \(2P = 14\)

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103.

The unit’s digit in the product \((3127)^{173}\)  is:

Answer: C

Unit digit in \((3127)^{173}\)  =  Unit digit in 7173. Now, 74 gives unit digit 1
Therefore, \(7173= (74)43 \times 71\). Thus, 7173 gives unit digit 7

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104.

Find the remainder when PP is divided by 9?

Answer: C

We need to find sum of S of the digits of PP.
Till 999 each of the digits from 1 to 9 occurs equal number of times.

Total digit of all the numbers from 1 to 999 will be multiple of ∑9=45.

Hence,

S=45k+(1+0+0+0)
⇒ Required remainder = 1.

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105.

Which digits should come in place of @ and # if the number 62684@# is divisible by both 8 and 5 ?

Answer: A

Since the given number is divisible by 5, so 0 or 5 must come in place of #.

But, a number ending with 5 is never divisible by 8.

So, 0 will replace #.

Now, the number formed by the last three digits is 4@0, which becomes divisible by 8, if @ is replaced by 4.

Hence, digits in place of @ and # are 4 and 0 respectively.

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106.

If each of the three nonzero numbers  a, b and  c  is divisible by 3, then abc must be divisible by which by ?

Answer: B

Since each one of the three numbers aa, bb, and cc is divisible by 3, the numbers can be represented as 3p,3qand 3r respectively, where p,q and r are integers.

The product of the three numbers is 3p×3q×3r=27(pqr).
Since p,qp,q and rr are integers, pqrpqr is an integer and therefore abcabc is divisible by 27.

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107.

A number when divided by 3 leaves a remainder 1. When the quotient is divided by 2, it leaves a remainder 1. What will be the remainder when the number is divided by 6?

Answer: C

Let \(n=3q+1\) and let \(q= 2p+1\). Then, \(n= 3(2p+1)+1= 6p+4\)
Therefore, the number when divided by \(6\), we get remainder\(= 4\)

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108.

The smallest prime number is?

Answer: C

A prime has exactly two distinct divisors, 1 and itself.

1 has only one divisor, and 0 has infinitely many, so neither qualifies. The smallest prime is 2 - and the only even one.

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109.

The sum of the present ages of two persons A and B is 60. If the age of A is twice that of B, find the sum of their ages 5 years hence?

Answer: C

\(A + B = 60, A = 2B\)
\(2B + B = 60 => B = 20\) then \(A = 40\)
5 years, their ages will be 45 and 25
Sum of their ages \(= 45 + 25 = 70\)

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110.

Which of the following is the minimum value of the sum of two integers whose product is 36?

Answer: D

List all possible factors xx and yy whose product is 36, and calculate the corresponding sum x+yx+y:

xx yy x×yx×y x+yx+y
1 36 36 37
2 18 36 20
3 12 36 15
4 9 36 13
5 6 36 12

From the table, the minimum sum is 12

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