Numbers with a given remainder

The idea

Numbers that leave remainder r when divided by n are r more than multiples of n, of the form nk + r; they can also be written as n – r less than multiples of n.

Remainder 3 on division by 5

The numbers 3, 8, 13, 18, 23 all leave remainder 3 when divided by 5.

Is 53 one of these numbers too?

• Yes
• No
• We cannot tell

Rows of five

Change k. Each number is k full rows of 5 and a last row with 3 cells. Look at the yellow cells that are missing from that last row.

r more, or n − r less

Numbers that leave remainder r when divided by n are r more than multiples of n, of the form nk + r; they can also be written as n – r less than multiples of n.

Remainder 3 on division by 5: 5k + 3 for k = 0 to 4: 3, 8, 13, 18, 23 5k − 2 for k = 1 to 5: 3, 8, 13, 18, 23

3 cells in the last row means 2 cells are missing: 5 − 3 = 2. So 3 more than a multiple of 5 is the same as 2 less than the next multiple of 5.

Notes

Numbers that leave remainder r when divided by n are r more than multiples of n, of the form nk + r; they can also be written as n – r less than multiples of n.

Check yourself

A number leaves remainder 4 when divided by 7. It has the form 7k + 4. What is the number when k = 6?

Answer: 46

7 × 6 + 4 = 42 + 4 = 46. And 46 = 6 × 7 + 4 leaves remainder 4 on division by 7.

Which number leaves remainder 2 when divided by 7?

A number is 2 less than a multiple of 5. What remainder does it leave when divided by 5?

Answer: 3

A multiple of 5 minus 2, such as 8 = 10 − 2, has 5 − 2 = 3 cells in its last row. The remainder is 3.

Which expression does not give numbers that leave remainder 4 when divided by 9? (k = 1, 2, 3 …)

• 15. 15 = 7 × 2 + 1. It leaves remainder 1.
• 23 — correct. Yes! 23 = 7 × 3 + 2.
• 31. 31 = 7 × 4 + 3. It leaves remainder 3.
• 41. 41 = 7 × 5 + 6. It leaves remainder 6.
• 9k + 4. k = 1 gives 13 = 9 + 4: remainder 4. This one is fine.
• 9k − 4 — correct. Yes! 9k − 4 is 4 less than a multiple of 9, so it leaves 9 − 4 = 5. (For k = 1: 5.)
• 9k − 5. k = 1 gives 4, k = 2 gives 13: remainder 4. 9k − 5 is 5 less than a multiple of 9, so it leaves 9 − 5 = 4. This one is fine.
• 9(k + 1) − 5. k = 1 gives 13 = 9 + 4: remainder 4. This one is fine.