Divisibility Rules & Remainder Theorem Questions and Answers

Answer: C

For a number to be divisible by 4, the number formed by its last two digits must be divisible by 4. The last two digits form the number '*4'.

We need to find the values of * such that '*4' is divisible by 4. Let's check single digits for *:

04 is divisible by 4 (so * can be 0).

14 is not divisible by 4.

24 is divisible by 4 (so * can be 2).

34 is not divisible by 4.

44 is divisible by 4 (so * can be 4).

54 is not divisible by 4.

64 is divisible by 4 (so * can be 6).

74 is not divisible by 4.

84 is divisible by 4 (so * can be 8).

The possible values for * are 0, 2, 4, 6, 8. The sum of these values is \(0+2+4+6+8 = 20\).

Answer: A

Correct answer: 3

Reduce each factor modulo 7 before multiplying, rather than forming the product first.

\[121 = 7\times 17 + 2,\qquad 122 = 7\times 17 + 3,\qquad 123 = 7\times 17 + 4\]

So the product is congruent to \(2\times 3\times 4 = 24\), and \(24 = 7\times 3 + 3\).

\[121\times 122\times 123 \equiv 3 \pmod 7\]

Answer: A

We find the remainder of each number when divided by 7 individually and then multiply the remainders.

Remainder of \(121 \div 7\) is 2 (since \(121 = 17 \times 7 + 2\)).

Remainder of \(122 \div 7\) is 3 (since \(122 = 17 \times 7 + 3\)).

Remainder of \(124 \div 7\) is 5 (since \(124 = 17 \times 7 + 5\)).

The remainder of the overall product is the remainder of \((2 \times 3 \times 5) \div 7\).

\(2 \times 3 \times 5 = 30\).

Now, we find the remainder of \(30 \div 7\).

\(30 = 4 \times 7 + 2\).

The remainder is 2.

Answer: D

We can solve this in two parts. First, find the remainder of \(9^{19} \div 8\). Since \(9 = 8+1\), the expression is \((8+1)^{19} \div 8\). The remainder will be \(1^{19}\), which is 1. Second, find the remainder of \(6 \div 8\), which is 6. The final remainder is the sum of these remainders, divided by 8 if necessary. Remainder = \((1 + 6) \div 8 = 7 \div 8\). The remainder is 7.

Answer: A

The largest 4-digit number is 9999. To find the largest 4-digit number divisible by 88, we divide 9999 by 88 and find the remainder.

\(9999 \div 88\).

\(9999 = 113 \times 88 + 55\).

The remainder is 55. This means 9999 is 55 more than a multiple of 88. To get the required number, we subtract the remainder from 9999.

Required Number = \(9999 - 55 = 9944\).

Answer: B

Let's analyze the expression in two parts modulo 68. Part 1: \(67^{67} \pmod{68}\). We can write 67 as (68 - 1). So, \((68-1)^{67} \pmod{68}\). This is equivalent to \((-1)^{67} \pmod{68}\), which is -1. Part 2: \(67 \pmod{68}\) is 67. Adding the remainders: \(-1 + 67 = 66\). The remainder is 66.

Answer: D

The dividend is a number formed by repeating the block '5812974' twice. Let X = 5812974. The dividend is \(X \times 10^7 + X = X(10^7 + 1)\). The quotient is X and the remainder is 0. According to the division algorithm, Dividend = Divisor × Quotient + Remainder. So, \(X(10^7+1) = Divisor \times X + 0\). Dividing by X, we get Divisor = \(10^7 + 1 = 10000001\).

Answer: A

For a number to be divisible by 99, it must be divisible by both 9 and 11.

Let's check option A: 114345.

1. Divisibility by 9: Sum of digits = \(1+1+4+3+4+5 = 18\). Since 18 is divisible by 9, the number is divisible by 9.

2. Divisibility by 11: Sum of alternate digits are \((5+3+1)=9\) and \((4+4+1)=9\). The difference is \(9-9=0\). Since the difference is 0, the number is divisible by 11.

Since 114345 is divisible by both 9 and 11, it is divisible by 99.

Answer: C

We find the cyclicity of remainders of powers of 3 when divided by 5.

\(3^1 \div 5 \rightarrow rem=3\)

\(3^2 \div 5 \rightarrow rem=4\) (from 9)

\(3^3 \div 5 \rightarrow rem=2\) (from 27)

\(3^4 \div 5 \rightarrow rem=1\) (from 81)

The cycle length is 4. We find the remainder of the power \(21\) when divided by 4, which is \(21 \div 4 \rightarrow rem=1\). The required remainder is the 1st in the cycle, which is 3.

Answer: A

Correct answer: 17

Successive division means the quotient from the first division is what gets divided next. Write \(N = 4x + 1\), where \(x\) is the first quotient, and then \(x = 5y + 4\).

The least value of \(N\) comes from the smallest admissible \(y\), which is \(y = 0\). That gives \(x = 4\) and

\[N = 4\times 4 + 1 = 17\]

Checking against the conditions: \(17 \div 4\) gives quotient 4 and remainder 1, and dividing that quotient, \(4 \div 5\) gives quotient 0 and remainder 4. Both requirements hold, so 17 is the least such number.