Divisibility Rules & Remainder Theorem Questions and Answers

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: \((y+z) \div x = (7+9) \div 4 = 16 \div 4 = 4\).

Answer: C

Let's use the property of remainders. We can find the remainder of each part separately. First part: \(29^{29} \div 30\). We can write 29 as \(30-1\). So we need the remainder of \((30-1)^{29} \div 30\). The remainder is \((-1)^{29} = -1\). Second part: \(29 \div 30\). The remainder is 29. Now, we add the remainders: \(-1 + 29 = 28\). The final remainder is 28.

Answer: A

Here, the difference between the divisor and the remainder is constant in each case: \(20-14=6\), \(25-19=6\), \(35-29=6\), \(40-34=6\). The required number is of the form (LCM of divisors) - (constant difference). First, find the LCM of 20, 25, 35, 40. \(20 = 2^2 \times 5\), \(25=5^2\), \(35=5 \times 7\), \(40=2^3 \times 5\). LCM = \(2^3 \times 5^2 \times 7 = 8 \times 25 \times 7 = 200 \times 7 = 1400\). The required number is \(1400 - 6 = 1394\).

Answer: A

Correct answer: 4

Reduce the base first: \(32 \equiv 5 \pmod 9\), so the problem becomes \(5^{32^{32}} \bmod 9\).

The powers of 5 modulo 9 run in a cycle of length 6:

\[5,\ 7,\ 8,\ 4,\ 2,\ 1,\ 5,\ 7,\ \ldots\]

So only the exponent modulo 6 matters. Since \(32 \equiv 2 \pmod 6\), we need \(2^{32} \bmod 6\). Powers of 2 alternate between 2 and 4 modulo 6, and an even exponent gives 4, so \(32^{32} \equiv 4 \pmod 6\).

The fourth entry of the cycle is 4, so \(32^{32^{32}} \equiv 5^4 = 625 \equiv 4 \pmod 9\).