Problem
Possible remainder values
A number should have been divided by 40 but was divided by 4. The wrong division's quotient and remainder are the right one's, swapped. A remainder is always smaller than the divisor. Find the smallest such number.
Reasoning · Grade 3-2 Completing Equations
Your answer
Watch out The quotient looks free to grow, but it doubles as the remainder after dividing by 4, so it must stay under 4.
How to solve
Strategy Convert to Algebra — The number can be written two ways with the same q and r appearing swapped, so writing both division equations and setting them equal pins down the relationship between q and r. The remainder-must-be-smaller rule limits q to just a few values, and the smallest valid one gives the smallest number, which I then verify directly.
1STEP 1
Write both divisions
With quotient q and remainder r the number is 40q + r and also 4r + q.
40q + r = 4r + q
Every division can be rebuilt as (divisor × quotient) + remainder, so I can write the one number two ways and compare.
4.NBT.B.6Convert To AlgebraAny division can be written back as divisor times quotient plus remainder, so one number can be described two ways and compared.
Why?
Splitting a number into whole equal groups plus a leftover is always possible and always in exactly one way.
🧱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?
The groups and the leftover fill the number with no gap and no overlap, so putting them back together rebuilds it exactly.
🧱Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.
2STEP 2
Find how q and r are related
Setting them equal gives r = 13q.
39q = 3r → r = 13q
Balancing the equation turns a wordy 'swap' condition into a tidy rule: the remainder is exactly 13 of the quotient.
4.OA.B.4Convert To Algebra3STEP 3
Use the remainder rules to limit q
Since q is a remainder mod 4, q is at most 3.
q ＜ 4, r = 13q ＜ 40 → q ∈ {1,2,3}
A leftover can never reach the size of the number you divide by, and that single rule shrinks the choices to just a handful.
4.NBT.B.6Guess And Check4STEP 4
Build the smallest number and check it
The smallest q = 1 gives r = 13, so the number is 53.
40 × 1 + 13 = 53; 53 ÷ 40 = 1 r13, 53 ÷ 4 = 13 r1
The smallest quotient gives the smallest number, and dividing it both ways confirms the quotient and remainder truly trade places.
4.NBT.B.6Guess And CheckAnswer
53
53 ÷ 40 = 1 r 13, 53 ÷ 4 = 13 r 1
53 is a sensible size: dividing by 40 once leaves 13, and dividing the same 53 by 4 leaves 1, and indeed 1 and 13 are swapped. Both remainders obey their rules (13 ＜ 40 and 1 ＜ 4). The next solutions would be q = 2 (number 106) and q = 3 (number 159), all larger, confirming 53 is the smallest.
Takeaway
Rebuild each number as (divisor × quotient) + remainder, set the two versions equal, and the 'smaller than the divisor' rule does most of the work for you!
- Write both divisions
- Find how q and r are related
- Use the remainder rules to limit q
- Build the smallest number and check it