+100 The sum of the prime numbers from 1-10, 11-20, 21-30, 31-40, and 41-50 are multiplied together with 997. What is the remainder when this result is divided by 97?
+100 1. Find the sums of the prime numbers in each interval
Primes from 1–10:
2, 3, 5, 7
Sum: (2 + 3 + 5 + 7 = 17)
Primes from 11–20:
11, 13, 17, 19
Sum: (11 + 13 + 17 + 19 = 60)
Primes from 21–30:
23, 29
Sum: (23 + 29 = 52)
Primes from 31–40:
31, 37
Sum: (31 + 37 = 68)
Primes from 41–50:
41, 43, 47
Sum: (41 + 43 + 47 = 131)
2. We want:
\( [ P = 17 \times 60 \times 52 \times 68 \times 131 \times 997 ] \), and we need the remainder when P is divided by 97.
3. Reduce each number modulo 97
17 is already < 97, so 17 mod 97 = 17
60 mod 97 = 60
52 mod 97 = 52
68 mod 97 = 68
131 mod 97 = 131 - 97 = 34
997 ÷ 97 = 10 remainder 27, so 997 mod 97 = 27
So:
\([ P \equiv 17 \times 60 \times 52 \times 68 \times 34 \times 27 \pmod{97} ]\).
4. Compute the product step by step mod 97.
Start multiplying step by step:
Calculate each step modulo 97.
Step 1: 17 × 60 = 1020
1020 ÷ 97 = 10 × 97 = 970, remainder 50
So 1020 mod 97 = 50
Step 2: 50 × 52 = 2600
2600 ÷ 97 = 26 × 97 = 2522, remainder 78
So 2600 mod 97 = 78
Step 3: 78 × 68 = 5304
5304 ÷ 97 = 54 × 97 = 5238, remainder 66
So 5304 mod 97 = 66
Step 4: 66 × 34 = 2244
2244 ÷ 97 = 23 × 97 = 2231, remainder 13
So 2244 mod 97 = 13
Step 5: 13 × 27 = 351
351 ÷ 97 = 3 × 97 = 291, remainder 60
So 351 mod 97 = 60.
Final answer: \(\boxed{60}\)
