UPSC CSE Prelims 2021
The divisibility rule for 7 states that if a number is partitioned into blocks of three digits from the right, and the alternating sum of these blocks is divisible by 7, then the original number is divisible by 7.
For the given number 3798125P369, the blocks of three digits, starting from the right, are 369, 25P, 981, and 37.
The alternating sum is calculated as: (369 - 25P + 981 - 37).
Simplifying the expression:
For the number 3798125P369 to be divisible by 7, the expression (1313 - 25P) must be divisible by 7.
First, determine the remainder of 1313 when divided by 7:
1313 = 7 × 187 + 4. Therefore, 1313 ≡ 4 (mod 7).
For (1313 - 25P) to be divisible by 7, it implies that (4 - 25P) must be divisible by 7. This condition is satisfied if 25P has a remainder of 4 when divided by 7.
The digit P can be any integer from 0 to 9. The number 25P represents a three-digit number where the hundreds digit is 2, the tens digit is 5, and the units digit is P.
Let us test the value P = 6 from the options:
Since 25P ≡ 4 (mod 7), the expression (4 - 25P) ≡ (4 - 4) ≡ 0 (mod 7). This confirms that (1313 - 25P) is divisible by 7 when P = 6.
The condition for divisibility by 7 requires that 25P must have a remainder of 4 when divided by 7. Let us examine the other options: