# Compatible numbers

Compatible numbers are numbers that seem to "go together" because they are easy to compute mentally. For various reasons, it is easier to compute addition, subtraction, multiplication, and division problems mentally using compatible numbers than numbers that are not compatible. As such, compatible numbers are commonly used to estimate the solutions to problems where some inaccuracy is acceptable.

One general rule that applies to all the basic arithmetic operations is that numbers that end in 0 (or multiple 0s) are compatible numbers, since in most cases they are easier to work with.

Examples of compatible numbers in addition include:

• 1 and 9
• 2 and 8
• 3 and 7
• 4 and 6
• 5 and 5

This includes all numbers that end in those pairs of numbers as well. This is because adding those specific numbers results in a number than ends in 0, which allows us to easily compute the addition mentally.

Examples

Find the sum of the following pairs of compatible numbers.

1. 21 + 39:

This can be broken up into 20 + 30 = 50 and 1 + 9 = 10, so

50 + 10 = 60

2. 433 + 127:

This can be broken up into 400 + 100 = 500, 30 + 20 = 50, and 7 + 3 = 10.

500 + 50 + 10 = 560

It is not actually necessary to break the numbers up if you recognize them, but when doing the mental math, recognizing these numbers is as if you are doing so. This is why such numbers are referred to as compatible numbers, since, with some practice, they become easily recognizable to us.

We can also fairly easily add numbers that end in 5 and 0. If there is an even number of 5s, the result will end in 0. If there is an odd number of 5s the result will end in 5. There are other compatible numbers as well, but these are some of the most common.

## Subtraction

Like in addition, numbers that end in 0 are compatible numbers because we can largely ignore the 0s, perform the subtraction, then attach the appropriate number of 0s after. Numbers that end with the same final digit are therefore compatible, since subtracting any number by itself results in 0.

Examples

Find the difference between the following pairs of compatible numbers.

1. 53 - 33:

This can be broken down into 50 - 30 (which can further be calculated as 5 - 3) = 20, and 3 - 3 = 0. Thus,

50 - 30 - 0 = 20

2. 1,500,000 - 12,000:

In this problem, the shared 0s can be dropped, leaving us with

1500 - 12 = 1488

Now we can attach the three shared 0s to the end to get the final answer: 1,488,000.

## Multiplication and division

Compatible numbers for multiplication and division are very similar. Like with addition and subtraction, numbers that end in one or more 0s are typically compatible. For multiplication, 0 multiplied by anything is 0, so zeros on the end of multiplication problems can be removed. For multiplication, any zeros at the end of the pairs of numbers can be ignored to simplify the problem, then added after the problem is complete. On the other hand, shared zeros in division problems can be canceled out.

Assuming that we know the multiplication table, all pairs in the multiplication table, including any variation with added 0s on the end, are compatible numbers.

Examples

Find the product or quotient of the following pairs of compatible numbers.

1. 40 × 3000:

40 × 3000 is the same as 4 × 3, with 4 added 0s.

40 × 3000 = 120,000

2. 810 ÷ 90:

810 and 90 have one shared 0, so we can cancel the 0s to get a division problem we should be familiar with if we've memorized the multiplication table.

81 ÷ 9 = 9

## Compatible numbers and estimation

Compatible numbers are often used to estimate values by converting actual values that we want to do arithmetic with to compatible numbers. Shopping is an activity in which we may use compatible numbers to get a rough estimate of how much we are going to spend before actually bringing our items to the register.

Example

Travis picks up 2 candy bars for \$0.99 each, an apple for \$1.69, a bag of chips for \$1.39, and a sandwich for \$4.88. Estimate how much Travis will have to pay for all of it together if there is a 10% sales tax.

First convert the prices involved to compatible numbers:

• \$0.99 is close to \$1, so 2 candy bars is ~\$2
• \$1.69 is close to 1.7 and 1.39 is close to 1.3, so the bag of chips and apple are ~\$3
• The sandwich is close to \$5.00
• 2 + 3 + 5 = \$10

10% of \$10 is the same as multiplying by 1/10, or dividing by 10, so 10% of \$10 is \$1. The total amount that Travis has to pay is therefore ~\$11. The exact amount is \$10.93, so our estimate is very close.

Estimating in this manner works for rough estimates, and in this case the calculations were simple enough that our estimate was close. However, if there are more values, and they are more complicated, it is important to try to balance how we estimate the values. For example, if in one case we estimate 1.6 as 2, we should try to keep track of this so we can round some other number down a similar amount. This way our estimate will be more accurate.