Facts for two, five and ten

When you reach for it

A child who knows what 8 \times 2 means, 8 equal groups of 2 as in What multiplication means, can always find it by counting. The aim now is to stop counting and know the product. The facts with a factor of 2, 5 or 10 are the place to start, because each one sits at the end of a skip count the child already says out loud:

• twos, from counting by twos: 2, 4, 6, 8, \dots
• tens, from counting by tens: 10, 20, 30, 40, \dots
• fives, from the same counting by fives: 5, 10, 15, 20, \dots

Reach for these facts whenever things come in twos, fives or tens: pairs of socks, shoes or mittens; nickels, which are worth 5 cents each; dimes, which are worth 10 cents each; fingers on a hand; tens in a number. Reach for them also as stepping stones. Later facts are built from these ones: 4 is two 2s, and 6 groups is 5 groups and one more, which is the work of Facts for three, four and six.

This lesson takes four days, one part each.

The procedure

Each part below gives one family of facts, says why the shortcut gives the right product, and shows the picture to draw. The facts run from a factor of 0 up to a factor of 10.

Part 1: twos are pairs and doubles

A pair of socks is 2 socks. So 8 pairs of socks is 8 groups of 2, and the number of socks is 8 \times 2. Count the socks by twos, one number for each pair, and stop when every pair has been counted.

The count stops at 16, so 8 \times 2 = 16. The count is right because each jump of two adds exactly one more pair, and there is one jump for each of the 8 pairs.

There is a second way to see a two fact. Put one sock from each pair in a left pile and the other sock in a right pile. Each pile has 8 socks, so the drawer holds 8 + 8 socks. That is a double:

So every two fact is a doubles fact, and a child who knows doubles already knows the twos. Either way, 2 \times 8 and 8 \times 2 give the same 16. The two orders are the same array of socks, 2 rows of 8 or 8 columns of 2; the commutative property says this always happens, so one fact covers both orders.

Every product with a factor of 2 is even. A pile of pairs can always be split into twos with nothing left over, which is what even means in Even and odd numbers. So the products of 2 end in 0, 2, 4, 6 or 8:

The same fact also answers the question run backward. If a drawer holds 14 socks in pairs, the number of pairs is the number in \square \times 2 = 14. Count by twos to 14 and count the jumps: 7 of them, so 7 pairs.

Part 2: tens are the digit with a zero

A dime is worth 10 cents. So 9 dimes are 9 groups of 10 cents, and their value is 9 \times 10 cents. Count them by tens.

The count stops at 90. There is a faster way to see it, from place value. Nine groups of ten are 9 tens, and 9 tens is written 90: a 9 in the tens place and a 0 in the ones place, because nothing is left over in the ones. So

The same reasoning works for every factor. n groups of ten are n tens, so n \times 10 is written as the digit n followed by a zero. At the top end, 10 \times 10 is ten tens, which is one hundred: 10 \times 10 = 100.

Every product of 10 ends in 0. A total like 45 cents cannot be made of dimes alone, since 4 dimes are 40 and 5 dimes are 50. Run backward, the tens digit gives the number of dimes: 80 cents is 8 tens, so it is 8 dimes.

Part 3: fives come from pairs of nickels

A nickel is worth 5 cents. So 7 nickels are worth 7 \times 5 cents. Counting by fives works: 5, 10, 15, 20, 25, 30, 35. The count is the slowest of the three, though, so here is a way to lean on the tens instead.

Two nickels are worth 5 + 5 = 10 cents, the same as one dime. So put the nickels in pairs. Each pair is worth 10 cents.

Seven nickels make 3 pairs, with 1 nickel left over. The 3 pairs are worth 3 tens, which is 30 cents, and the extra nickel adds 5:

This explains the pattern in the fives. When the number of nickels is even, they pair up exactly, so the product is a whole number of tens and ends in 0. When the number is odd, one nickel is left over, so the product ends in 5. Halve the even number (or the even number just below an odd one) and you have the tens digit.

Look at the table next to the tens table. Each five fact is half of the ten fact under the same n: 5 \times 6 = 30 and 10 \times 6 = 60. That is the same fact about coins said the other way round. A nickel is worth half a dime, so n nickels are worth half as much as n dimes.

Part 4: putting the three families together

A real handful holds more than one kind of thing, so the last part uses the three families side by side. A pile of nickels and dimes is counted one kind at a time: find the value of the nickels with a five fact, the value of the dimes with a ten fact, then add the two values. That is allowed because every coin belongs to exactly one of the two kinds, so nothing is counted twice and nothing is missed.

The families also check each other. A dime is worth two nickels, so any number of dimes trades for twice as many nickels with the same value. Five dimes are worth the same as ten nickels:

So 5 \times 10 and 10 \times 5 name the same 50 cents, which is the commutative property again, now seen with coins. And the ending digits give a quick check on any answer:

• a product of 2 is even;
• a product of 10 ends in 0;
• a product of 5 ends in 0 or 5.

An answer that breaks its rule is wrong. A product that keeps its rule may still be wrong, so the rule is a check, not a way to find the product. The patterns of the whole table, beyond what these three families need, wait for Patterns in the multiplication table.

Worked examples

Examples 1 and 2 go with Part 1, Examples 3 and 4 with Part 2, Examples 5 and 6 with Part 3, and Examples 7 and 8 with Part 4.

Boundaries

• Whole groups only. Each fact counts whole groups: whole pairs, whole coins. Half a pair or half a nickel is not a group of this kind, and products like that need fractions, which come later.
• Factors from 0 to 10. These are the facts up to 10 \times 10. Past ten the same counts keep going (11 \times 2 = 22, 12 \times 10 = 120), but those products are not facts to know by heart here.
• Zero groups give zero. 0 \times 2, 0 \times 5 and 0 \times 10 are all 0: no pairs is no socks, and no coins is no cents.
• The ending rules are checks, not answers. An answer of 16 for 2 \times 9 is even, so it passes the rule for twos, but 2 \times 9 is 18. The rule can show an answer is wrong; it cannot show an answer is right.
• A backward question needs an exact fit. \square \times 10 = 45 has no whole number answer, because 45 cents cannot be made of dimes alone. Likewise \square \times 2 is never an odd number.
• Same kind of thing. Socks add to socks and cents to cents. A pile of 3 nickels and 2 dimes is 15 + 20 = 35 cents. Adding the coins first, 3 + 2 = 5 coins, counts coins, not cents, because the two kinds of coin have different values.

Where this goes next

These three families are the first block of the multiplication table to be known by heart, and the next lessons lean on them. The properties of multiplication turn every fact into two, so 7 \times 5 also answers 5 \times 7. The The distributive property builds a hard fact from easy ones, and the easy ones are mostly these. Facts for three, four and six gets the fours by doubling the twos and the sixes from the fives plus one more group, and Facts for seven, eight and nine finishes the table the same way.

Each fact also answers a division question, as Examples 2 and 4 already did: \square \times 2 = 18 is the same question as 18 \div 2, and Division facts makes that link for the whole table. The ending-digit rules return in Patterns in the multiplication table, and much later as the divisibility rules for very large numbers. The tens facts are also the first step toward Multiplying by tens and hundreds, where 4 \times 30 is read as 4 \times 3 tens.

Exercises

Exercises 1 to 6 go with Part 1, 7 to 12 with Part 2, 13 to 16 with Part 3, and 17 to 21 with Part 4. The reasoning group draws on all four.

Part 1: twos

1. Find 2 \times 6 and 6 \times 2.
2. Vaiana’s drawer holds 5 pairs of socks. How many socks is that?
3. Find 7 \times 2 twice: once by counting by twos and once as a double.
4. Calum counts 20 socks, every one with its partner. How many pairs are there?
5. Fill in the boxes: 2 \times \square = 8 and \square \times 2 = 6.
6. Could a drawer of complete pairs hold 13 socks? Explain.

Part 2: tens

7. Find 10 \times 1, 4 \times 10 and 10 \times 10.
8. Tuuli has 8 dimes. How many cents does she have?
9. Calum has 20 cents, all in dimes. How many dimes does he have?
10. Fill in the box: 10 \times \square = 30.
11. Vaiana has 2 dimes and Tuuli has 5 dimes. How many cents do they have together?
12. Which of these amounts could be made of dimes alone: 30 cents, 45 cents, 100 cents, 72 cents?

Part 3: fives

13. Find 5 \times 4 and 5 \times 1.
14. Vaiana has 2 nickels. How many cents does she have?
15. Use pairs of nickels to find 5 \times 5.
16. Tuuli has 20 cents, all in nickels. How many nickels does she have?

Part 4: all three together

17. Vaiana has 4 nickels and Calum has 4 dimes. Who has more money, and how many cents more?
18. Tuuli has 3 nickels and 6 dimes. How many cents does she have?
19. Fill in the box: 5 \times 4 = 10 \times \square. Explain with coins.
20. Can 15 cents be paid with dimes only? With nickels only? Give the number of coins where it can.
21. Calum has 8 dimes and 1 nickel. How many cents does he have?

Reasoning

22. Explain why every product with a factor of 2 is even.
23. Explain, using place value, why 10 \times 7 = 70.
24. Explain with pairs of nickels why 5 times an even number ends in 0 and 5 times an odd number ends in 5.
25. Is 5 \times n always half of 10 \times n? Explain with nickels and dimes.