Question

The least number which when divided by 9 and 15 leaves remainder 7 in each case is

Answer 1

Answer:

52 is answer of this question

Answer 2

Answer:

Since the remainder is 7, there exist (nonzero) natural numbers k and p such that n=9p+7 and n=15k+7.

Since n=n, we have 9p+7=15k+7.

Subtract 7 from each side: 9p=15k

Divide both sides by 3: 3p=5k

Since 3 is a prime number that divides into the product 5k, either 3 divides into 5 or 3 divides into k. Since 3 doesn’t divide into 5, we have that 3 divides into k.

The smallest positive number that is divisible by 3 is 3. If k=3, then n=15*3+7=52. If n<52, then k<3, which we just showed cannot be true.

Therefore, the smallest number that is larger than 9 or 15 and leaves a remainder of 7 when divided by 9 or 15 is 52.

Step-by-step explanation:

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