ऐसे कितने धन पूर्णांक हैं जिसे 1186 को विभाजित करने पर शेषफल 31 आता है ?
सही उत्तर (C) 8 है।
व्याख्या:
When 1186 is divided by a natural number n, it leaves a remainder of 31. This means that (1186 - 31) is completely divisible by n. Let us calculate 1186 - 31 = 1155. Therefore, n must be a divisor (factor) of 1155. Also, by the definition of division and remainders, the divisor n must be strictly greater than the remainder. Since the remainder is 31, we must have: n > 31. Now, let us find all the factors of 1155 and count how many of them are greater than 31. First, find the prime factorization of 1155: 1155 / 5 = 231 231 / 3 = 77 77 / 7 = 11 11 / 11 = 1 So, prime factorization of 1155 = 3 × 5 × 7 × 11. The total number of divisors of 1155 is (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 × 2 = 16 divisors. Let us list the divisors of 1155 and identify those that are greater than 31: All divisors of 1155: 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155. Let us check which divisors are > 31: - 33 > 31 (1) - 35 > 31 (2) - 55 > 31 (3) - 77 > 31 (4) - 105 > 31 (5) - 165 > 31 (6) - 231 > 31 (7) - 385 > 31 (8) - 1155 > 31 (9)? Wait! Let us count carefully. Let us list all divisors greater than 31: 33, 35, 55, 77, 105, 165, 231, 385, 1155. That is 9 divisors! Wait, let us re-verify the options: 6, 7, 8, 9. Option (d) is 9. Wait, let us check if any divisor <= 31 gives remainder 31. No, divisor must be > 31. Let us re-verify the list of divisors of 1155: Divisors: 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155. Total 16 divisors. Divisors <= 31 are: 1, 3, 5, 7, 11, 15, 21 (7 divisors). Total divisors (16) - Divisors <= 31 (7) = 9 divisors! Thus, exactly 9 natural numbers satisfy the condition. Option (d) is correct.
In English (Question & Model Answer)
How many natural numbers are there which give a remainder of 31 when 1186 is divided by these natural numbers ?
The correct answer is (C) 8.
Explanation:
When 1186 is divided by a natural number n, it leaves a remainder of 31. This means that (1186 - 31) is completely divisible by n. Let us calculate 1186 - 31 = 1155. Therefore, n must be a divisor (factor) of 1155. Also, by the definition of division and remainders, the divisor n must be strictly greater than the remainder. Since the remainder is 31, we must have: n > 31. Now, let us find all the factors of 1155 and count how many of them are greater than 31. First, find the prime factorization of 1155: 1155 / 5 = 231 231 / 3 = 77 77 / 7 = 11 11 / 11 = 1 So, prime factorization of 1155 = 3 × 5 × 7 × 11. The total number of divisors of 1155 is (1 + 1)(1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 × 2 = 16 divisors. Let us list the divisors of 1155 and identify those that are greater than 31: All divisors of 1155: 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155. Let us check which divisors are > 31: - 33 > 31 (1) - 35 > 31 (2) - 55 > 31 (3) - 77 > 31 (4) - 105 > 31 (5) - 165 > 31 (6) - 231 > 31 (7) - 385 > 31 (8) - 1155 > 31 (9)? Wait! Let us count carefully. Let us list all divisors greater than 31: 33, 35, 55, 77, 105, 165, 231, 385, 1155. That is 9 divisors! Wait, let us re-verify the options: 6, 7, 8, 9. Option (d) is 9. Wait, let us check if any divisor <= 31 gives remainder 31. No, divisor must be > 31. Let us re-verify the list of divisors of 1155: Divisors: 1, 3, 5, 7, 11, 15, 21, 33, 35, 55, 77, 105, 165, 231, 385, 1155. Total 16 divisors. Divisors <= 31 are: 1, 3, 5, 7, 11, 15, 21 (7 divisors). Total divisors (16) - Divisors <= 31 (7) = 9 divisors! Thus, exactly 9 natural numbers satisfy the condition. Option (d) is correct.