A number is divided by 4; the quotient is then divided by 5; and that quotient is divided by 6. The three remainders come out as 2, 3 and 4 in that order. What is the smallest such number?

Rebuild the number from the last division upwards. The final quotient is allowed to be 0, so the second quotient is $6\times 0+4=4$, the first quotient is $5\times 4+3=23$, and the number is $4\times 23+2=94$. Check: 94 ÷ 4 = 23 remainder 2, then 23 ÷ 5 = 4 remainder 3, then 4 ÷ 6 = 0 remainder 4. The trap is 214 — it satisfies all three conditions as well, but it is the NEXT one up (94 + 120), not the smallest. Both 50 and 122 break down at the second division.