Problem
Find the divisor from remainder clues
Dividing 50 by a mystery number leaves 4. Dividing 120 by the same number leaves 5. Several numbers may behave this way. Find the biggest one.
Reasoning · Grade 4-1 Division
Your answer
Watch out The common factor 1 looks eligible, but it is smaller than the remainder 5.
How to solve
Strategy Make a Systematic List — Remainders are awkward to search with directly, so I work backwards from what a remainder means: if 50 leaves 4 behind, then taking those 4 away leaves a number the mystery number divides exactly. That turns two messy remainder clues into two clean divides-exactly clues. Then I make a systematic list of every factor of each of those two numbers, keep only the factors on both lists, and eliminate the ones too small to leave a remainder of 5.
1STEP 1
Turn the remainders into exact divisions
Removing the remainders leaves 46 and 115.
50 - 4 = 46, 120 - 5 = 115
Working backwards from the leftovers is like handing back the extra sweets before sharing: whatever is left in the bowl now shares out perfectly.
4.NBT.B.6Work Backwards2STEP 2
List every factor of 46 and of 115
46 has factors 1, 2, 23 and 46.
46 = 1 × 46 = 2 × 23, 115 = 1 × 115 = 5 × 23
Hunting for factor pairs in order — 1, then 2, then 3, and so on — guarantees none is missed, and each pair found also hands over its partner for free.
4.OA.B.4Make A Systematic List3STEP 3
Keep only the numbers on both lists
Both lists share 1 and 23.
{1, 2, 23, 46} and {1, 5, 23, 115} → {1, 23}
Only two short lists have to be compared, so matching them up is easier than testing divisors one by one all the way to 46.
4.OA.B.4Make A Systematic List4STEP 4
Throw out the candidate that is too small
1 is smaller than the remainder 5, so it drops out.
1 ＜ 5, so 1 cannot leave remainder 5
If the leftover were as big as the divisor, another whole group could still be handed out, so remainders are always the smaller number — an easy test that kills 1 straight away.
4.NBT.B.6Eliminate PossibilitiesA divisor of 1 is impossible, because a remainder must always be smaller than the number you divide by.
Why?
If the leftover were as big as the divisor, one more whole group could still be handed out, so the leftover stops just below it.
🧱Division with remainderAny number splits into a whole count of equal groups plus a leftover, and the leftover is always smaller than the group size.
Why?
That single test kills the smallest candidate at once, leaving only one number on both lists to check properly.
🧱Rule out the rest and the last one is forcedIf every possibility but one has been ruled out, the survivor must be the answer — no further checking is needed.
5STEP 5
Check 23 against both clues
The number that fits both clues is 23.
50 = 23 × 2 + 4, 120 = 23 × 5 + 5
Rebuilding each dividend as quotient times divisor plus remainder is the standard way to check a division, and it confirms the answer without any new ideas.
4.NBT.B.6Eliminate PossibilitiesAnswer
23
50 = 23 × 2 + 4, 120 = 23 × 5 + 5
The answer is a whole number bigger than 5, as it must be to leave a remainder of 5, and smaller than 46, as it must be to divide 46. Both original statements check out exactly: 23 times 2 plus 4 is 50, and 23 times 5 plus 5 is 120. It also makes sense that only one number worked — 46 and 115 share just the factor 23 apart from 1, so there was never room for a bigger answer.
Takeaway
Take the leftovers away first — then the messy remainder puzzle turns into an ordinary hunt for common factors!
- Turn the remainders into exact divisions
- List every factor of 46 and of 115
- Keep only the numbers on both lists
- Throw out the candidate that is too small
- Check 23 against both clues