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.

Q: When 1721 is divided by 16 then remainder is :

The remainder is 1.Since 17 = 16 + 1, 17 leaves remainder 1 when divided by 16. Every power of 17 does the same, because remainders multiply.\[ 17^{21} = (16 + 1)^{21} \equiv 1^{21} = 1 \pmod{16} \]In the binomial expansion of \((16 + 1)^{21}\), every term except the last 1 contains a factor 16. So the remainder cannot be 0, 2 or 3.

Important Instructions

31.

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.

Discuss in Forum

32.

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.

Discuss in Forum

33.

When 17²¹ is divided by 16 then remainder is :

Exam Year(s)

UKPSC Lower PCS - 2025,

Answer: D

The remainder is 1.

Since 17 = 16 + 1, 17 leaves remainder 1 when divided by 16. Every power of 17 does the same, because remainders multiply.

\[ 17^{21} = (16 + 1)^{21} \equiv 1^{21} = 1 \pmod{16} \]

In the binomial expansion of \((16 + 1)^{21}\), every term except the last 1 contains a factor 16. So the remainder cannot be 0, 2 or 3.

Discuss in Forum

34.

Which of the following numbers is exactly divisible by 3?

A.23456
B.54321
C.78901
D.13579

Answer: B

A number is divisible by 3 if the sum of its digits is a multiple of 3.

A) 2+3+4+5+6 = 20 (Not divisible by 3)

B) 5+4+3+2+1 = 15 (Divisible by 3, since 15 = 3 × 5)

C) 7+8+9+0+1 = 25 (Not divisible by 3)

D) 1+3+5+7+9 = 25 (Not divisible by 3)

Discuss in Forum

35.

What is the remainder when 1234 is divided by 5?

Answer: D

To find the remainder when a number is divided by 5, we only need to look at its last digit. The remainder is the same as the remainder of the last digit when divided by 5.

The last digit is 4. When 4 is divided by 5, the remainder is 4.

Discuss in Forum

36.

A number is divisible by 8 if the number formed by its last three digits is divisible by 8. Which of the following numbers is divisible by 8?

A.56780
B.12345
C.98764
D.45672

Answer: D

We check the number formed by the last three digits for divisibility by 8.

A) 780 ÷ 8 = 97.5 (Not divisible)

B) 345 ÷ 8 = 43.125 (Not divisible)

C) 764 ÷ 8 = 95.5 (Not divisible)

D) 672 ÷ 8 = 84 (Divisible). So, 45672 is divisible by 8.

Discuss in Forum

37.

What least value must be assigned to '*' so that the number 63576*2 is divisible by 8?

Answer: C

For a number to be divisible by 8, the number formed by its last three digits must be divisible by 8. Here, the number is 6*2.

Let's check values for *:

If * = 1, we have 612. \(612 \div 8 = 76.5\) (Not divisible)

If * = 2, we have 622. \(622 \div 8 = 77.75\) (Not divisible)

If * = 3, we have 632. \(632 \div 8 = 79\) (Divisible)

The least value is 3.

Discuss in Forum

38.

What is the remainder when \(17^{200}\) is divided by 18?

A.1
B.2
C.16
D.17

Answer: A

We can use the remainder theorem. \(17\) can be written as \((18 - 1)\). So, we need to find the remainder of \((18-1)^{200} \div 18\).

Using the binomial expansion, every term except the last one will have 18 as a factor. The last term is \((-1)^{200}\).

\((-1)^{200} = 1\).

So the remainder is 1.

Discuss in Forum

39.

Which digit should replace '*' in the number 987*54 to make it divisible by 6?

Answer: B

For a number to be divisible by 6, it must be divisible by both 2 and 3.

1. Divisibility by 2: The number ends in 4, which is even. So, it is divisible by 2.

2. Divisibility by 3: The sum of its digits must be divisible by 3. Sum = \(9+8+7+*+5+4 = 33 + *\).

We need \(33 + *\) to be a multiple of 3. Since 33 is already a multiple of 3, * must also be a multiple of 3. The possible single-digit values are 0, 3, 6, 9. From the options, the value is 3.

Discuss in Forum

40.

Find the remainder when \(2^{31}\) is divided by 5.

Answer: C

We look at the pattern of remainders of powers of 2 when divided by 5.

\(2^1 \div 5 \rightarrow rem=2\)

\(2^2 \div 5 \rightarrow rem=4\)

\(2^3 \div 5 \rightarrow rem=3\)

\(2^4 \div 5 \rightarrow rem=1\)

The cycle of remainders is (2, 4, 3, 1), which has a length of 4. To find the remainder of \(2^{31}\), we find the remainder of the power when divided by the cycle length: \(31 \div 4 \rightarrow rem=3\). The required remainder is the 3rd in the cycle, which is 3.

Discuss in Forum

Loading…

Topics

Other Subjects

Tags: Divisibility Rules Questions and Answers || Remainder Theorem MCQ Questions and Answers || Divisibility and Remainder Theorem GK Questions || GK on Divisibility Rules || Quantitative Aptitude Questions and Answers || Divisibility Theorem MCQ

Divisibility Rules & Remainder Theorem Questions and Answers

Important Instructions

Topics

Other Subjects

Tags: Divisibility Rules Questions and Answers || Remainder Theorem MCQ Questions and Answers || Divisibility and Remainder Theorem GK Questions || GK on Divisibility Rules || Quantitative Aptitude Questions and Answers || Divisibility Theorem MCQ

About Us

Stay informed about all important Dates & Notifications