Divisibility Rules & Remainder Theorem Questions and Answers

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.

Answer: C

Correct answer: 5

Since \(70 = 7 \times 10\), the number must be divisible by both factors.

Divisibility by 10 fixes the last digit at once: \(y = 0\). The number is then \(73100 + 10x\).

\[73100 = 7\times 10442 + 6 \quad\Longrightarrow\quad 73100 \equiv 6 \pmod 7\]

We need \(6 + 10x \equiv 0 \pmod 7\). As \(10 \equiv 3 \pmod 7\), this is \(3x \equiv 1 \pmod 7\); multiplying by 5, the inverse of 3 modulo 7, gives \(x \equiv 5\). The only digit satisfying that is \(x = 5\).

The number is 73150, and \(73150 \div 70 = 1045\). Hence \(x - y = 5 - 0 = 5\).

Answer: D

For a number to be divisible by 90, it must be divisible by both 9 and 10.

1. Divisibility by 10: The last digit must be 0. So, \(y=0\).

2. Divisibility by 9: The sum of the digits must be divisible by 9. The number is now 732x0. The sum of digits is \(7+3+2+x+0 = 12+x\).

For \(12+x\) to be divisible by 9, the sum must be the next multiple of 9, which is 18. So, \(12+x = 18\), which gives \(x=6\).

We have \(x=6\) and \(y=0\). Therefore, \(x+y = 6+0 = 6\).

Answer: D

Correct answer: 25

Rather than expanding \(4^{19}\), find how soon the powers of 4 repeat modulo 33.

\[4^1 = 4,\quad 4^2 = 16,\quad 4^3 = 64 \equiv 31,\quad 4^4 \equiv 124 \equiv 25,\quad 4^5 \equiv 100 \equiv 1 \pmod{33}\]

The cycle length is 5. Since \(19 = 5\times 3 + 4\),

\[4^{19} = \left(4^5\right)^3 \times 4^4 \equiv 1^3 \times 25 = 25 \pmod{33}\]

Answer: B

We can use Euler's Totient Theorem. The totient of 33, \(\phi(33) = \phi(3 \times 11) = \phi(3) \times \phi(11) = (3-1)(11-1) = 2 \times 10 = 20\).

According to Euler's theorem, if a and n are coprime, then \(a^{\phi(n)} \equiv 1 \pmod{n}\). Here, 4 and 33 are coprime.

So, \(4^{20} \equiv 1 \pmod{33}\).

We need to find the remainder of \(4^{21} \div 33\).

\(4^{21} = 4^{20} \times 4^1\).

So, \(4^{21} \pmod{33} \equiv (4^{20} \pmod{33}) \times (4^1 \pmod{33}) \pmod{33}\).

\(\equiv 1 \times 4 \pmod{33}\).

The remainder is 4.

Answer: C

Correct answer: 6

Since \(72 = 8 \times 9\) and 8 and 9 share no factor, both tests must pass.

Divisibility by 8. Only the last three digits matter: 63y, that is \(630 + y\). Because \(630 = 8\times 78 + 6\), we need \(6 + y \equiv 0 \pmod 8\), and the only digit that works is \(y = 2\).

Divisibility by 9. The digit sum must be a multiple of 9: \(5+7+x+6+3+2 = 23 + x\). The only digit giving a multiple of 9 is \(x = 4\), for a sum of 27.

The number is 574632, and \(574632 \div 72 = 7981\). Hence \(x + y = 4 + 2 = 6\).

Answer: D

Correct answer: 7

Since \(72 = 8 \times 9\), apply both tests.

Divisibility by 8. The last three digits are 83y, that is \(830 + y\). Because \(830 = 8\times 103 + 6\), we need \(6 + y \equiv 0 \pmod 8\), giving \(y = 2\) as the only digit.

Divisibility by 9. The digit sum is \(4+1+x+4+8+3+2 = 22 + x\), which must be a multiple of 9. Only \(x = 5\) qualifies, for a sum of 27.

The number is 4154832, and \(4154832 \div 72 = 57706\). Hence \(x + y = 5 + 2 = 7\).

Answer: B

A number that is divisible by 7, 11, and 13 is also divisible by their product, which is \(7 \times 11 \times 13 = 1001\). Any 6-digit number formed by repeating a 3-digit number, i.e., of the form 'abcabc', is divisible by 1001. This is because 'abcabc' = abc × 1001. In our case, the number is 479xyz. For it to be divisible by 1001, it must be of the form 479479. By comparing, we get x=4, y=7, and z=9. Now, we calculate the required value: \((y+z) \div x = (7+9) \div 4 = 16 \div 4 = 4\).