An expert question on remainders for P5–P6, with a step-by-step solution, diagrams and the answer. Common in the Hua Xia Cup's P3 to P6 papers; harder ones need several remainder conditions to hold at once.
The question
A puzzle on the maths noticeboard:
"I am a whole number.
Divided by 23, my quotient equals my remainder; divided by 29, my quotient equals my remainder too.
Neither remainder is 0."
What do all the whole numbers that fit add up to?
Show the hints (3)
💡Hint 1: Dividing by 23, call the quotient and the remainder both k: how can the number be written with k? How big can k be?
💡Hint 2: The number = 24 × k with k from 1 to 22; likewise it is a multiple of 30. How big can each condition let the number be?
💡Hint 3: The number is a common multiple of 24 and 30, and it must meet both limits.
Worked solution
Have a go first, then read the solution step by step.
1.Divided by 23 with quotient and remainder both k: the number = 23 × k + k = 24 × k. The remainder is below 23 and not 0, so k runs 1 to 22: a multiple of 24, at most 24 × 22 = 528.
2.Likewise for 29: a multiple of 30, at most 30 × 28 = 840. Both limits apply, so at most 528.
3.The LCM of 24 and 30 is 120, so the number is a multiple of 120: those up to 528 are 120, 240, 360, 480 — 4 of them.
⚠️Common mistake: using only the looser limit 840 and counting 600 and the like, getting 3360; but 600 ÷ 23 gives a quotient bigger than the remainder. Others take the common multiple as 24 × 30 = 720.
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