What is the positive integer n that leaves a remainder of 2 when divided by 3 and a remainder of 1 when divided by 5?

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To find the positive integer n that satisfies the given conditions, we can represent the problem using modular arithmetic.

1. The first condition states: when n is divided by 3, the remainder is 2. In mathematical terms, we can express this as:

n ≡ 2 (mod 3)

2. The second condition states: when n is divided by 5, the remainder is 1. We express this as:

n ≡ 1 (mod 5)

Now we have the two modular equations:

  • n ≡ 2 (mod 3)
  • n ≡ 1 (mod 5)

To solve these congruences, we can find a common solution by checking values that meet the first condition and see if they satisfy the second:

Starting with the first condition, possible values of n can be represented as:

  • n = 3k + 2 for integer k

Now we can check different values of k:

  • If k = 0, then n = 3(0) + 2 = 2
  • If k = 1, then n = 3(1) + 2 = 5
  • If k = 2, then n = 3(2) + 2 = 8
  • If k = 3, then n = 3(3) + 2 = 11
  • If k = 4, then n = 3(4) + 2 = 14
  • If k = 5, then n = 3(5) + 2 = 17

Now each of these values should be checked to find one that meets the second condition:

  • n = 2: 2 mod 5 = 2 (not valid)
  • n = 5: 5 mod 5 = 0 (not valid)
  • n = 8: 8 mod 5 = 3 (not valid)
  • n = 11: 11 mod 5 = 1 (valid)
  • n = 14: 14 mod 5 = 4 (not valid)
  • n = 17: 17 mod 5 = 2 (not valid)

We found that when n = 11, it satisfies both conditions:

  • When 11 is divided by 3, the remainder is 2.
  • When 11 is divided by 5, the remainder is 1.

Therefore, the positive integer n that meets both conditions is 11.

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