Divisibility Rules & Remainder Theorem Questions and Answers

Q: Which of the numbers given below is that number which is divisible by all numbers from 1 to 9 except 5 and 7 ?

The number is 144.It must be divisible by 1, 2, 3, 4, 6, 8 and 9, so it must be a multiple of their least common multiple:\[ \begin{aligned} \text{LCM}(8, 9) &= 72 \\ 144 &= 2 \times 72 \end{aligned} \]Since 8 and 9 already cover 2, 3, 4 and 6, any multiple of 72 works, and 144 is the only one offered. 140 is not divisible by 3 or 8, 142 is not divisible by 3 or 4, and 148 is not divisible by 3 or 8.

Q: The value of 351 mod 5 is __________.

351 mod 5 is 1.The mod operation gives the remainder after division: 351 = 5 × 70 + 1, so the remainder is 1. A quick check is that a number's remainder on division by 5 depends only on its last digit, and 1 leaves remainder 1.0 would need the number to end in 0 or 5. A remainder on division by 5 is always between 0 and 4, so 5 and 8 are impossible.In programming terms this is 351 % 5, which also evaluates to 1.

Important Instructions

121.

Which of the numbers given below is that number which is divisible by all numbers from 1 to 9 except 5 and 7 ?

A.140
B.142
C.144
D.148

Exam Year(s)

Tripura Public Service Commission - 2014,

Answer: C

The number is 144.

It must be divisible by 1, 2, 3, 4, 6, 8 and 9, so it must be a multiple of their least common multiple:

\[ \begin{aligned} \text{LCM}(8, 9) &= 72 \\ 144 &= 2 \times 72 \end{aligned} \]

Since 8 and 9 already cover 2, 3, 4 and 6, any multiple of 72 works, and 144 is the only one offered. 140 is not divisible by 3 or 8, 142 is not divisible by 3 or 4, and 148 is not divisible by 3 or 8.

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122.

The value of 351 mod 5 is __________.

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: B

351 mod 5 is 1.

The mod operation gives the remainder after division: 351 = 5 × 70 + 1, so the remainder is 1. A quick check is that a number's remainder on division by 5 depends only on its last digit, and 1 leaves remainder 1.

0 would need the number to end in 0 or 5. A remainder on division by 5 is always between 0 and 4, so 5 and 8 are impossible.

In programming terms this is 351 % 5, which also evaluates to 1.

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