Numbers and Divisibility

To test divisibility by 11, calculate the alternating sum of digits: (1-4+3-1+4-3+1-4+3-1+4-3) = 0. Since the result is 0 (a multiple of 11), the number is divisible by 11. The pattern 143 repeats, and 143 itself equals 11×13, making any repetition divisible by 11.

To check divisibility by 17, divide 124362769 by 17. The result is 7315457, which is an integer with no remainder, so the number is divisible by 17. When one integer divides another evenly (remainder = 0), the second number is a factor of the first.

Need n² + n - 90 ≡ 0 (mod 17), which means (2n+1)² ≡ 361 ≡ 1 (mod 17), so 2n+1 ≡ ±1 (mod 17). This gives n ≡ 0 or 12 (mod 17). In [10,100]: n = 12, 29, 46, 63, 80, 97 (for n ≡ 12) and n = 17, 34, 51, 68, 85 (for n ≡ 0). Total = 10 values.

Using Legendre's formula for exponent of 2 in 12!: floor(12/2)=6, floor(12/4)=3, floor(12/8)=1, floor(12/16)=0. Sum = 10. So 2¹⁰ divides 12! completely, leaving an odd quotient (all 2s removed).

132 = 4 × 3 × 11. A number is divisible by 132 only if divisible by all three. Checking: 264 (132×2, yes), 396 (132×3, yes), 462 (not by 4, no), 792 (132×6, yes), 968 (not by 3, no), 2178 (not by 4, no), 5184 (not by 11, no), 6336 (132×48, yes). Exactly 4 numbers are divisible.

To be divisible by 132, a number must be divisible by 3, 4, and 11. Testing the numbers reveals that exactly 4 of them (264, 396, 792, and 6336) are divisible by 132.

11158 divided by 77 gives remainder 70. To make it divisible, we need to add 77 - 70 = 7. Checking: 11158 + 7 = 11165, and 11165 / 77 = 145 exactly. Options A, B, and D would give remainders of 76, 77, and 75 respectively, not divisibility.

We can't judge the no's in 2's complement first we need to convert them in decimal Given no. $ 11111011 \rightarrow 00000101 = 5 $
$ 11100111 \rightarrow 00011001 = 25 $ $ 11100100 \rightarrow 00011100 = 28 $
$ 11010111 \rightarrow 00101001 = 41 $ $ 11011011 \rightarrow 00100101 = 37 $

From all only option (1) is divisible by 5. Shortcut: To convert 2's complement no. directly into original binary, we should complement all the digits from MSB till the last one (1). Keep the last 1 from the LSB as it is. Observe the example.

We can't judge the no's in 2's complement first we need to convert them in decimal Given no. $ 11111011 \rightarrow 00000101 = 5 $
$ 11100111 \rightarrow 00011001 = 25 $ $ 11100100 \rightarrow 00011100 = 28 $
$ 11010111 \rightarrow 00101001 = 41 $ $ 11011011 \rightarrow 00100101 = 37 $

From all only option (1) is divisible by 5. Shortcut: To convert 2's complement no. directly into original binary, we should complement all the digits from MSB till the last one (1). Keep the last 1 from the LSB as it is. Observe the example.

We can't judge the no's in 2's complement first we need to convert them in decimal Given no. $ 11111011 \rightarrow 00000101 = 5 $
$ 11100111 \rightarrow 00011001 = 25 $ $ 11100100 \rightarrow 00011100 = 28 $
$ 11010111 \rightarrow 00101001 = 41 $ $ 11011011 \rightarrow 00100101 = 37 $

From all only option (1) is divisible by 5. Shortcut: To convert 2's complement no. directly into original binary, we should complement all the digits from MSB till the last one (1). Keep the last 1 from the LSB as it is. Observe the example.

The least number to be added to the numbers to make them divisible by 9 is equal to the difference of the least multiple of 9 greater than the sum of the digits and sum of the digits. Sum of the digits = 15. Nearest multiple of 9 greater than the sum of the digits = 18. Hence, 3 has to be added. If we add 3 to 51234, we get 51237, which is fully divisible by 9.

305678956 is divisible by 2 because it is an even number and all even numbers are divisible by 2.