Fill the bags
There are 37 sweets, and they go into bags of 5. Fill one bag, then the next, and keep going until you run out of sweets.
Seven bags fill up, and they hold 7 × 5 = 35 sweets. That leaves 2 sweets. They go into an eighth bag, but 2 sweets cannot fill a bag of 5, so the eighth bag stays short by 3.
Seven full bags of 5, and a last bag with 2 sweets in it and 3 empty places.
The same thing as a division
Putting 37 sweets into bags of 5 is the division 37 ÷ 5. The number of full bags is the answer, and the sweets left over are the remainder: 37 ÷ 5 = 7 remainder 2, which is also written 7 r 2.
Work it out the way long division does. 5 goes into 37 seven times, because 7 × 5 = 35 and 8 × 5 = 40 is too many. Subtract to find what is left: 37 − 35 = 2.
Multiplying back checks it. Seven bags of 5 and 2 more make 7 × 5 + 2 = 35 + 2 = 37, which is every sweet.
7 × 5 = 35 is written under the 37, and 37 − 35 = 2 is the remainder.
Always smaller than the divisor
The 2 sweets are the remainder because they are too few to fill another bag. If 5 or more were left over, another bag of 5 would fill, and the division would not be finished.
So when you divide by 5, the remainder is always 0, 1, 2, 3 or 4. A remainder of 5 or more means the answer above it is too small: go back and fit in another 5.
37 = 5 × 7 + 2: 7 full bags and 2 left over, fewer than 5, so another bag still needs 3: 2 + 3 = 5
Add items one at a time until the remainder resets
Add sweets one at a time. Each one goes into the short bag, and the remainder goes up: 2, then 3, then 4. The next sweet fills the bag, so there is one more full bag and the remainder goes back to 0. It never reaches 5.
Two slips
The remainder is not the number of bags. For 37 ÷ 5, the 7 counts the full bags and the 2 counts the sweets left over. When a question asks for the remainder, the answer is 2.
The remainder is not the number still needed to fill the next bag either. That is 3, the gap from 37 up to 40. The remainder is what is left over from 37 after 35, and 37 − 35 = 2.
Worked example: Periodic / Cyclic Remainder Pattern
Question A string of decorative beads is arranged in a repeating pattern: 2 Red, 3 Blue, 1 Yellow, 2 Green, followed by 2 Red, 3 Blue, 1 Yellow, 2 Green, and so on. (a) What is the color of the 147th bead? (b) How many Blue beads are there in the first 200 beads?
1.Draw 1 Unit Block = [R][R][B][B][B][Y][G][G] (8 beads total, containing 3 Blue).
One block: 2 red, 3 blue, 1 yellow, 2 green. Eight beads, three of them blue. 2.(a) Find number of blocks in 147 beads: 147 ÷ 8 = 18 blocks with a remainder of 3 beads.
(a) 147 ÷ 8 = 18 blocks with 3 over. 3.Trace first 3 beads of the next block: [R], [R], [B] ⟹ 3rd bead is Blue.
Count 3 into the next block: blue. Bead 147 is blue. 4.(b) Number of complete blocks in 200 beads: 200 ÷ 8 = 25 complete blocks.
(b) 200 beads is 200 ÷ 8 = 25 whole blocks. 5.Each block contains 3 Blue beads: 25 × 3 = 75 Blue beads.
Each block has 3 blue: 25 × 3 = 75.
Answer: (a) Blue; (b) 75 Blue beads
Common mistakes
- Misinterpreting a remainder of 0 as the 0th bead instead of the very last bead of the cycle (Green).
- Miscounting the cumulative positions within the cycle (e.g., assuming Blue begins at position 2 instead of position 3).
More patterns and page numbers problems, worked step by step →