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) n is odd n ≡ 1 (mod 5) n ≡ 0 (mod 7) n < 100
Step 1: Numbers ≡ 1 (mod 15): 1, 16, 31, 46, 61, 76, 91
Step 2: Check which are divisible by 7: Only 91 =
Step 3: Verify 91 is odd and less than 100
Step 4: Verify all conditions: 91 ÷ 3 = 30 R1 91 ÷ 5 = 18 R1 91 ÷ 7 = 13 R0
Answer: The number of pebbles is 91.
Explanation
The riddle gives us these conditions: (1) When divided by 3, remainder is 1, so n ≡ 1 (mod 3). (2) When divided by 2 (pairing), remainder is 1, meaning n is odd. (3) When divided by 5, remainder is 1, so n ≡ 1 (mod 5). (4) When divided by 7, remainder is 0, so n ≡ 0 (mod 7). (5) Less than 100. Numbers ≡ 1 (mod 15) are: 1, 16, 31, 46, 61, 76, 91. Of these, only 91 is divisible by 7. Verification: 91 ÷ 3 = 30 remainder 1, 91 is odd, 91 ÷ 5 = 18 remainder 1, 91 ÷ 7 = 13 remainder 0.
Solution Steps
Step 1: From the riddle: n ≡ 1 (mod 3), n is odd, n ≡ 1 (mod 5), n ≡ 0 (mod 7), n < 100
Step 2: Numbers ≡ 1 (mod 15): 1, 16, 31, 46, 61, 76, 91
Step 3: Check which are divisible by 7: Only 91 =
Step 4: Verify 91 is odd and less than 100
Step 5: Verify all conditions: 91 ÷ 3 = 30 R1, 91 ÷ 5 = 18 R1, 91 ÷ 7 = 13 R0