A fruit-seller has a basket of mangoes. Counted out in threes, 2 are left over; counted out in fives, 3 are left over; counted out in sevens, 2 are left over. What is the smallest number of mangoes the basket can hold?

The threes and the sevens leave the SAME remainder, 2, so the count is 2 more than a multiple of 21: the candidates are 23, 44, 65, 86, 107, … . Test each against the fives, and 23 works first time — indeed $23=3\times 7+2=5\times 4+3=7\times 3+2$. The trap 128 meets all three conditions as well, but it is the NEXT one up, 23 + 105, where $105=3\times 5\times 7$ is the gap between consecutive solutions. Meanwhile 17 fails on the fives $(17=5\times 3+2)$ and 53 fails on the sevens $(53=7\times 7+4)$, each satisfying two conditions out of three.