Question 5
“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?”

We need to find a number of pebbles that fits all the given conditions.
Step 1 — Write down the conditions
Let us call the number of pebbles N.
The riddle gives us several clues about N.

Step 2 — Find a general form for N
We see that N leaves a remainder of 1 when divided by 2, 3, and 5. This means that if we subtract 1 from N, the new number must be perfectly divisible by 2, 3, and 5. So, is a common multiple of 2, 3, and 5. We find the Least Common Multiple (LCM) of these numbers.
So, must be a multiple of 30. We can write this as:
Step 3 — Test values for N
Now we use the other two conditions: N must be a multiple of 7, and N must be less than 100. Let us try different values for k in the equation .
If :
1 is not a multiple of 7.
If :
31 is not a multiple of 7 ( with remainder 3).
If :
61 is not a multiple of 7 ( with remainder 5).
If :
91 is a multiple of 7 (). Also, 91 is less than 100. This value of N satisfies all the conditions.
If :
121 is greater than 100, so it does not fit the last condition. Therefore, we stop here.
The only number that satisfies all the conditions is 91.
Answer
The number of pebbles is 91.
More questions in FIO
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(i) The sum of two even numbers is a multiple of 3.
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(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
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“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?”
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(ii) 4779 - 661
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