# 33 and Some Divisibility Tricks for 3 and 9

33 is a composite number. 33 = 1 x 33 or 3 x 11. Factors of 33: 1, 3, 11, 33. Prime factorization: 33 = 3 x 11. When 33 is a clue in the FIND THE FACTORS 1 – 12 puzzles, always use 3 and 11 as the factors.

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What patterns do you see in the following chart? Probably you noticed or a teacher taught you the easy way to remember what 9 times a numbers from one to ten is just as the chart illustrates.

Did you also ever notice that if you add up the digits of the multiples of 9 in the multiplication table, you get 9?

What is really great is if you add up the digits of ANY multiple of 9, you’ll get 9 or some other multiple of 9! This is called a divisibility trick because it is a way to find out if a number is divisible by 9 without actually dividing by 9.

The same trick works on multiples of 3: If you add up the digits of a multiple of 3, you will get 3 or some other multiple of 3! Lets apply these divisibility tricks to a few numbers:

Is 243 divisible by 3 or 9? We don’t have to divide to know the answer:

2 + 4 + 3 = 9, which is divisible by both 3 and 9, so, yes, 243 is divisible by both 3 and 9.

If you do the actual division:

• 243 ÷ 3 = 81
• 243 ÷ 9 = 27.

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Is 582 divisible by 3 or by 9? Add up the digits to find out: 5 + 8 + 2 = 15.

Since 15 is divisible by 3, but not by 9, we know 582 is divisible by 3, but not by 9.

If you do the actual division:

• 582 ÷ 3 = 194
• 582 ÷ 9 = 64 Remainder 6.

When we added the digits of 582, we got 15. Notice that 1 + 5 = 6, the remainder when we divided 582 by 9.

When you add up the digits of a number until you have only one digit left, if that digit is not 9, then that number is the remainder you would get if you did the actual division!

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Now let’s see if 1753 is divisible by 3 or 9.

1 + 7 + 5 + 3 = 16; 1 + 6 = 7.

7 cannot be evenly divided by 3 or 9, so 1753 is not divisible by 3 or 9.

If you do the actual division:

• 1753 ÷ 9 = 194 Remainder 7 (the same 7 that equals 1 + 6 above).
• 1753 ÷ 3 = 584 Remainder 1

Notice that 7 ÷ 3 = 2 Remainder 1

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All of these problems demonstrate that if you add the digits of a number until you are left with a single digit, if that digit is 3, 6, or 9, then the original number is divisible by 3.

The last problem demonstrates that if you divide that single digit by 3, the remainder will be the same if you divided the original number by 3.

These divisibility tricks for 3 and 9 can give quite a bit of valuable information!

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