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Patterns in the multiplication table

Patterns and sequences · level 3, lesson 5 of 6

A question to start

Write down the last digit of each answer in the 4 times table: 4, 8, 12, 16, 20, 24, 28, ... gives

4, 8, 2, 6, 0, 4, 8, 2, 6, 0, ...

Now draw a circle with the digits 0 to 9 around it like a clock, and join the digits in that order. You get a five-pointed star. Do the same with the 6 times table and you get the same star, drawn the other way round.

0 1 2 3 4 5 6 7 8 9 4s: 4, 8, 2, 6, 0, ... 0 1 2 3 4 5 6 7 8 9 6s: 6, 2, 8, 4, 0, ...

Why would two different times tables draw the same star? The multiplication table is full of patterns like this, and every one of them has a reason you can find.

The idea

The table itself

Here is the multiplication table up to \(10 \times 10\). Each answer sits where its row and column meet: row 3, column 5 holds \(3 \times 5 = 15\). The square numbers are in bold.

× 1 2 3 4 5 6 7 8 9 10
1 1 2 3 4 5 6 7 8 9 10
2 2 4 6 8 10 12 14 16 18 20
3 3 6 9 12 15 18 21 24 27 30
4 4 8 12 16 20 24 28 32 36 40
5 5 10 15 20 25 30 35 40 45 50
6 6 12 18 24 30 36 42 48 54 60
7 7 14 21 28 35 42 49 56 63 70
8 8 16 24 32 40 48 56 64 72 80
9 9 18 27 36 45 54 63 72 81 90
10 10 20 30 40 50 60 70 80 90 100

The square numbers run down the diagonal

The diagonal is the slanting line of boxes from the top-left corner to the bottom-right: row 1 column 1, row 2 column 2, and so on. Each of those boxes multiplies a number by itself, so the diagonal holds the square numbers 1, 4, 9, 16, ..., 100 from lesson 2.

The table is the same when you flip it

Find 15 in row 3, column 5. Now look in row 5, column 3: also 15. In fact, if you folded the table along the diagonal, every number would land on an equal number. Why?

3 rows of 5 5 rows of 3 turn a quarter turn same 15 dots: 3 × 5 = 5 × 3

\(3 \times 5\) means 3 rows of 5 dots. Turn that picture a quarter turn and it becomes 5 rows of 3 dots, which is \(5 \times 3\). Turning does not add or remove any dots, so the two answers must be equal. That is true for any pair of numbers, so the whole table is symmetric about its diagonal: the top-right half is a mirror copy of the bottom-left half. It also means you only have to learn about half the table; the other half comes free.

Why the last digits of the 4s repeat

The last digits of the 4 times table go 4, 8, 2, 6, 0 and then start again. Here is the reason.

The 5th answer is \(4 \times 5 = 20\), which ends in 0. Now compare any answer with the one five places later: \(4 \times 6\) is \(4 \times 1\) plus five more fours, which is \(4 + 20 = 24\). \(4 \times 7\) is \(4 \times 2 + 20 = 28\). Adding 20 adds 2 to the tens but leaves the ones digit alone. So every five places, the last digit comes back to what it was. The pattern must repeat every 5.

The 6s work the same way, because \(6 \times 5 = 30\) also ends in 0: 6, 2, 8, 4, 0, then again.

Why the 4s and the 6s draw the same star backwards

Put the two lists side by side:

× 1 2 3 4 5
4s end in 4 8 2 6 0
6s end in 6 2 8 4 0
sum of the two 10 10 10 10 0

Each pair of last digits adds to 10 (or both are 0). The reason: four of something plus six of the same thing is ten of it. \(4 \times 3 + 6 \times 3 = 10 \times 3 = 30\), which ends in 0. So the last digits of \(4 \times 3\) and \(6 \times 3\) must add up to 10 (or both be 0). On the clock of digits, a digit and its partner that adds to 10 sit in mirror places (4 and 6, 8 and 2), so the 6s visit the same five points as the 4s, in the opposite direction.

The same idea pairs other tables: 2 and 8 (\(2 + 8 = 10\)), and 3 and 7 (\(3 + 7 = 10\)).

The 9s: up one, down one

9, 18, 27, 36, 45, 54, 63, 72, 81, 90. The tens digit goes up by 1 each time and the ones digit goes down by 1, and the two digits always add to 9. Why? Adding 9 is the same as adding 10 and then taking away 1. Adding 10 puts 1 on the tens; taking away 1 takes 1 off the ones.

Hiding next to the diagonal

Look at the boxes just beside the diagonal: \(1 \times 2 = 2\), \(2 \times 3 = 6\), \(3 \times 4 = 12\), \(4 \times 5 = 20\), \(5 \times 6 = 30\). Halve them: 1, 3, 6, 10, 15. The triangular numbers from lesson 3! That is exactly the rectangle trick: a rectangle 4 by 5 is two staircases of 10.

And look across the diagonal from 25: the boxes \(4 \times 6 = 24\), \(3 \times 7 = 21\), \(2 \times 8 = 16\), \(1 \times 9 = 9\). They are \(25 - 1\), \(25 - 4\), \(25 - 9\), \(25 - 16\): the square minus the square numbers! To see one of these: start from a 5 by 5 square (25 dots), take off one column of 5 to leave 5 rows of 4 (20 dots), then add one more row of 4 to get 6 rows of 4 (24 dots). You took off 5 and put back 4, so you are 1 short: \(24 = 25 - 1\).

Worked examples

1. The last digit of \(4 \times 37\), without multiplying it all out. The last digits of the 4s repeat every 5. \(37 \div 5 = 7\) remainder 2, so \(4 \times 37\) ends like \(4 \times 2 = 8\): it ends in 8. (Check: \(4 \times 37 = 148\). ✓)

2. The last digit of \(6 \times 23\). The 6s also repeat every 5. \(23 \div 5 = 4\) remainder 3, so it ends like \(6 \times 3 = 18\): in 8. (Check: \(6 \times 23 = 138\). ✓)

3. \(9 \times 7\) by the up-one-down-one rule. Counting down the 9 times table: 9, 18, 27, 36, 45, 54, 63. The tens digit is one less than 7, which is 6, and the ones digit makes the digits add to 9: 3. So 63.

4. Around 49. \(7 \times 7 = 49\). Then, going across the diagonal: \(6 \times 8 = 49 - 1 = 48\), \(5 \times 9 = 49 - 4 = 45\), \(4 \times 10 = 49 - 9 = 40\). All correct.

Try it

  1. Write the last digits of the 8 times table for \(8 \times 1\) to \(8 \times 10\). How often do they repeat?
  2. Do the last digits of the 3 times table use all ten digits before they repeat?
  3. What is the last digit of \(4 \times 58\)? Use the repeating pattern, then check by multiplying.
  4. \(10 \times 10 = 100\). Without multiplying in the usual way, find \(9 \times 11\) and \(8 \times 12\).
  5. \(6 \times 6 = 36\). What is \(5 \times 7\)?
  6. \(5 \times 6 = 30\) sits just beside the diagonal. Halve it. Which triangular number is it?
  7. Explain, with a picture of dots, why \(4 \times 7 = 7 \times 4\).
  8. The 3s end in 3, 6, 9, 2, 5, 8, 1, 4, 7, 0. Which other times table has the same last digits in the reverse order? Explain why.
Answers
  1. 8, 6, 4, 2, 0, 8, 6, 4, 2, 0. They repeat every 5, because \(8 \times 5 = 40\) ends in 0.
  2. Yes: 3, 6, 9, 2, 5, 8, 1, 4, 7, 0 uses every digit once. It only repeats after 10, when \(3 \times 10 = 30\) ends in 0.
  3. \(58 \div 5 = 11\) remainder 3, so it ends like \(4 \times 3 = 12\): in 2. Check: \(4 \times 58 = 232\). ✓
  4. Across the diagonal from 100: \(9 \times 11 = 100 - 1 = 99\) and \(8 \times 12 = 100 - 4 = 96\).
  5. \(36 - 1 = 35\).
  6. \(30 \div 2 = 15\), the 5th triangular number (\(1 + 2 + 3 + 4 + 5\)).
  7. Draw 4 rows of 7 dots. Turn the drawing a quarter turn: now it is 7 rows of 4 dots. It is the same dots, so the number is the same: 28.
  8. The 7s: 7, 4, 1, 8, 5, 2, 9, 6, 3, 0. Three of something plus seven of the same thing is ten of it, which ends in 0, so each pair of last digits adds to 10 (or both are 0): \(3 + 7\), \(6 + 4\), \(9 + 1\), and so on.

Common mistake

Thinking a pattern repeats with the wrong length. The 4s' last digits repeat every 5 answers, not every 4. The length comes from the first answer that ends in 0 (\(4 \times 5 = 20\)), not from the number 4 itself.

Reading only the last digit. The pattern 4, 8, 2, 6, 0 tells you the last digit, not the whole answer. \(4 \times 37\) ends in 8, but it is 148, not 8 or 28.

Where it leads