What remains

Find a number that leaves a remainder of 3 when divided by 5. There's 8, and 13, and 18, and 23, and 3 itself. Can one expression capture all of them?

Start from what you know. Numbers leaving remainder 0 on division by 5 are the multiples of 5, of the form 5k. (In general, number = divisor × quotient + remainder.)

Numbers leaving remainder 3 are 3 more than multiples of 5. So they are 5k + 3. Put k = 0, 1, 2, 3, 4 and out come 3, 8, 13, 18, 23: exactly the list.

There's a second description of the same set. Those numbers are also 2 less than multiples of 5: 5k − 2, with k ≥ 1. Put k = 1, 2, 3, 4, 5 and you get 3, 8, 13, 18, 23 again.

So two different-looking expressions can make the very same set. Check candidates carefully: 3k + 5, 3k − 5 and 5k − 3 all look plausible, and none of them works.

Now combine two conditions. Find numbers leaving remainder 2 on division by 3 and remainder 2 on division by 4.

If the number is x, then x − 2 is a multiple of 3 and also a multiple of 4, so x − 2 is a multiple of LCM(3, 4) = 12. Hence x = 12n + 2, giving 14, 26, 38, and so on. Peel off the remainder first; what's left must be a common multiple.

The book sets the idea as a riddle: '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.' (One more than a common multiple of 2, 3 and 5, so 31, 61 or 91; only 91 = 7 × 13 works.)