I hold some pebbles, not too many, When I group them in 3's, one stays with me. Try pairing them up — it simply won't do, A stubborn odd pebble remains in my view. Group them by 5, yet one's still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?
Given: From the riddle:
- n ≡ 1 (mod 3) — remainder 1 when divided by 3
- n is odd — cannot be paired
- n ≡ 1 (mod 5) — remainder 1 when divided by 5
- n ≡ 0 (mod 7) — divisible by 7
- n < 100
Solution: Step 1: From riddle: n ≡ 1 (mod 3), n is odd, n ≡ 1 (mod 5), n ≡ 0 (mod 7), n < 100
Step 2: From n ≡ 1 (mod 3) and n ≡ 1 (mod 5): n ≡ 1 (mod 15), so n = 15k + 1
Step 3: From n ≡ 0 (mod 7): 15k + 1 ≡ 0 (mod 7), so 15k ≡ 6 (mod 7)
Step 4: Since 15 ≡ 1 (mod 7): k ≡ 6 (mod 7), so k = 7m + 6
Step 5: n = 15(7m + 6) + 1 = 105m + 91
Step 6: For m = 0: n = 91; for m = 1: n = 196 > 100
Step 7: Verify: 91 ÷ 3 = 30 R1, 91 is odd, 91 ÷ 5 = 18 R1, 91 ÷ 7 = 13 R0
Answer: The number of pebbles is 91.
Explanation
From the riddle: (1) n ≡ 1 (mod 3) - remainder 1 when divided by 3, (2) n is odd - cannot be paired, (3) n ≡ 1 (mod 5) - remainder 1 when divided by 5, (4) n ≡ 0 (mod 7) - divisible by 7, (5) n < 100. From conditions 1 and 3: n ≡ 1 (mod 15), so n = 15k + 1. From condition 4: 15k + 1 ≡ 0 (mod 7), so 15k ≡ -1 ≡ 6 (mod 7). Since 15 ≡ 1 (mod 7), we get k ≡ 6 (mod 7). So k = 7m + 6, giving n = 15(7m + 6) + 1 = 105m + 91. For m = 0: n = 91. For m = 1: n = 196 > 100. So the answer is 91.
Solution Steps
Step 1: From riddle: n ≡ 1 (mod 3), n is odd, n ≡ 1 (mod 5), n ≡ 0 (mod 7), n < 100
Step 2: From n ≡ 1 (mod 3) and n ≡ 1 (mod 5): n ≡ 1 (mod 15), so n = 15k + 1
Step 3: From n ≡ 0 (mod 7): 15k + 1 ≡ 0 (mod 7), so 15k ≡ 6 (mod 7)
Step 4: Since 15 ≡ 1 (mod 7): k ≡ 6 (mod 7), so k = 7m + 6
Step 5: n = 15(7m + 6) + 1 = 105m + 91
Step 6: For m = 0: n = 91; for m = 1: n = 196 > 100
Step 7: Verify: 91 ÷ 3 = 30 R1, 91 is odd, 91 ÷ 5 = 18 R1, 91 ÷ 7 = 13 R0