Determine which of the given numbers divides the sum 7^15 + 7^16 + 7^17.

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We need to check divisibility of the sum 7^15 + 7^16 + 7^17 by 3, 5, 2, and 4.

Step 1: Check divisibility by 3.
- Consider 7 mod 3: 7 ≡ 1 (mod 3)
- Therefore, 7^n ≡ 1^n ≡ 1 (mod 3) for any integer n.
- So, 7^15 + 7^16 + 7^17 ≡ 1 + 1 + 1 = 3 ≡ 0 (mod 3).
- Hence, the sum is divisible by 3.

Step 2: Check divisibility by 5.
- 7 mod 5 = 2.
- Powers of 2 mod 5 cycle every 4: 2^1=2, 2^2=4, 2^3=3, 2^4=1.
- Find 7^15 mod 5 = 2^15 mod 5.
- 15 mod 4 = 3, so 2^15 mod 5 = 2^3 = 3.
- Similarly, 7^16 mod 5 = 2^16 mod 5 = 2^(16 mod 4) = 2^0 = 1.
- 7^17 mod 5 = 2^17 mod 5 = 2^(1) = 2.
- Sum mod 5 = 3 + 1 + 2 = 6 ≡ 1 (mod 5), not divisible by 5.

Step 3: Check divisibility by 2.
- 7 is odd, so any power of 7 is odd.
- Sum of three odd numbers is odd + odd + odd = odd.
- Odd number is not divisible by 2.

Step 4: Check divisibility by 4.
- 7 mod 4 = 3.
- Powers of 3 mod 4 cycle every 2: 3^1=3, 3^2=1.
- 7^15 mod 4 = 3^(15) mod 4.
- 15 mod 2 = 1, so 3^(15) mod 4 = 3.
- 7^16 mod 4 = 3^(16) mod 4.
- 16 mod 2 = 0, so 3^(16) mod 4 = 1.
- 7^17 mod 4 = 3^(17) mod 4.
- 17 mod 2 = 1, so 3^(17) mod 4 = 3.
- Sum mod 4 = 3 + 1 + 3 = 7 ≡ 3 (mod 4), not divisible by 4.

Conclusion: The sum is divisible only by 3 among the given options.

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