Divisibility Rules & Remainder Theorem Questions and Answers

Answer: A

For a number to be divisible by 8, the number formed by its last three digits must be divisible by 8. The number is 5x4. We need to find the smallest digit x such that 5x4 is divisible by 8. Let's test values for x starting from 0. If x=0, the number is 504. \(504 \div 8 = 63\). It is divisible. Therefore, the smallest integer value for x is 0.

Answer: C

We can use Fermat's Little Theorem. Since 7 is a prime number, \(3^{7-1} = 3^6 \equiv 1 \pmod{7}\). We need to find the remainder for \(3^{10}\). We can write \(3^{10} = 3^6 \times 3^4 \equiv 1 \times 3^4 \pmod{7}\). Now we just need to calculate \(3^4 = 81\). The remainder of \(81 \div 7\) is 4 (since \(81 = 11 \times 7 + 4\)). So the remainder is 4.

Answer: B

The required number is the HCF of the differences between the given numbers. The differences are \((132 - 62) = 70\), \((237 - 132) = 105\), and \((237 - 62) = 175\). We need to find the HCF of 70, 105, and 175. \(70 = 2 \times 5 \times 7\), \(105 = 3 \times 5 \times 7\), \(175 = 5^2 \times 7\). The highest common factor is \(5 \times 7 = 35\).

Answer: D

The expression \(n^2 - 1\) can be factored as \((n-1)(n+1)\). Since n is an odd integer, both \(n-1\) and \(n+1\) are consecutive even integers. For example, if n=3, we get 2 and 4. If n=5, we get 4 and 6. The product of two consecutive even integers is always divisible by 8. This is because one of them must be a multiple of 4. Let the consecutive even integers be 2k and 2k+2. Their product is \(4k(k+1)\). Since either k or k+1 must be even, \(k(k+1)\) is always divisible by 2. Thus the product is divisible by \(4 \times 2 = 8\).

Answer: C

For a number to be divisible by 24, it must be divisible by its co-prime factors, 3 and 8. Let's check the options.

A) 35718: Sum of digits is 24 (divisible by 3). Last 3 digits 718. \(718 \div 8\) is not an integer. Not divisible by 8.

B) 63810: Sum is 18 (divisible by 3). Last 3 digits 810. \(810 \div 8\) is not an integer. Not divisible by 8.

C) 53784: Sum is 27 (divisible by 3). Last 3 digits 784. \(784 \div 8 = 98\). Divisible by 8. Since it is divisible by both 3 and 8, it is divisible by 24.

Answer: A

We use Euler's Totient Theorem. First, find \(\phi(15) = \phi(3 \times 5) = \phi(3)\phi(5) = (3-1)(5-1) = 2 \times 4 = 8\). According to the theorem, \(2^8 \equiv 1 \pmod{15}\). We need to find the remainder of the power \(2024\) when divided by 8. The rule for divisibility by 8 is to check the last three digits. \(024\) is divisible by 8. So, \(2024\) is a multiple of 8. We can write \(2024 = 8k\). The expression becomes \((2^8)^k \equiv 1^k \equiv 1 \pmod{15}\). The remainder is 1.

Answer: C

For a number to be divisible by 72, it must be divisible by 8 and 9.

1. Divisibility by 8: The number formed by the last three digits, 73*, must be divisible by 8. Let's test values for *. 730 is not. 731 is not. 732 is not. 733 is not. 734 is not. 735 is not. 736 is. \(736 \div 8 = 92\). So, the missing digit * must be 6.

2. Divisibility by 9: Let's confirm with this value. The number is 425736. The sum of digits is \(4+2+5+7+3+6 = 27\). Since 27 is divisible by 9, the number is divisible by 9. Since it satisfies both conditions, the value of * is 6.

Answer: A

By the Polynomial Remainder Theorem, the remainder when a polynomial P(x) is divided by \(x-a\) is P(a). Here, P(x) = \(x^2 - 7x + 15\) and a=3. We calculate P(3). \(P(3) = (3)^2 - 7(3) + 15 = 9 - 21 + 15 = 24 - 21 = 3\). The remainder is 3.

Answer: D

Let the 2-digit number be 'ab'. The 4-digit number formed is 'abab'. This can be written mathematically as \(1000a + 100b + 10a + b = 1010a + 101b = 101(10a+b)\). Since \(10a+b\) represents the original 2-digit number, the 4-digit number 'abab' is always a multiple of 101. Therefore, it is always exactly divisible by 101.

Answer: A

We can write 397 as \(398-1\). The expression becomes \((398-1)^{3589} + 5 \pmod{398}\). The first term \((398-1)^{3589}\) has a remainder of \((-1)^{3589} = -1\) when divided by 398. The second term, 5, has a remainder of 5. The total remainder is \(-1 + 5 = 4\). The remainder is 4.