### 138 and Divisibility Tricks 4 You

138 is a composite number. Factor pairs: 138 = 1 x 138, 2 x 69, 3 x 46, or 6 x 23. Factors of 138: 1, 2, 3, 6, 23, 46, 69, 138. Prime factorization: 138 = 2 x 3 x 23.

138 is never a clue in the FIND THE FACTORS puzzles.

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After you learned some basic division facts, you probably realized:

- 2 will divide evenly into any EVEN whole number.
- 5 will divide evenly into whole numbers ending in 0 or 5.
- 10 will divide evenly into whole numbers ending in 0.

These three rules are related to each other. All of them are true because we use base ten in our numbering system, and the prime factorization of 10 is 2 x 5.

If you needed to find the factors of a 33-digit whole number, you would be able to tell if 2, 5, or 10 divide evenly into it just by looking at the last digit. 33-digits is more than a standard calculator can handle, but no matter how many digits a whole number has, as long as you can see the very last one, you can apply those three simple divisibility rules to know if 2, 5, or 10 are factors. Thus you will be able to do something a calculator can’t.

But wait, there are even more divisibility tricks if you can see the last TWO digits of the whole number!

- 10 squared, better known as 100, divides evenly into any whole number ending in 00.
- 5 x 5 = 25 which divides evenly into any whole number ending in 00, 25, 50, or 75.
- 2^2 (AKA 4) divides evenly into a whole number if the final two digits can be divided evenly by 4.

How can one tell if the last two digits of a whole number are divisible by 4 (without actually dividing by 4)? I’ll show you how: I’ve put the 25 possible 2-digit multiples of 4 into one of two lists:

- 00, 04, 08, 20, 24, 28, 40, 44, 48, 60, 64, 68, 80, 84, 88
- 12, 16, 32, 36, 52, 56, 72, 76, 92, 96

Notice in the first list ALL the digits are even and the last digit (0, 4, or 8) can be divided evenly by 4.

Then look at the second list. The first digit is always odd and the last digit is either 2 or 6 (the only two even digits that are not divisible by 4).

Hmm. I think we can rewrite the divisibility rule for 4:

- 4 (AKA 2^2) divides evenly into a whole number if the last two digits are even and the final digit is divisible by 4 (the last digit is 0, 4, or 8).
- 4 divides evenly into any whole number whose next to the last digit is odd if the final digit is even but not divisible by 4 (the last digit is 2 or 6).

The rewritten divisibility rule is longer to read but takes a little less time to implement so you will have to decide which version of the rule works best for you. Either trick takes much less time than dividing some really long whole number by 4 or dividing by 2 twice.

Now I’m on to thinking about what the last THREE digits tell us.

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