5 is one number that fits this description, but there are many others as well.

The way to answer a problem like this is to look for a pattern.

You know that if you divide the number by 2, there will be a remainder of 1. This eliminates half of...

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5 is one number that fits this description, but there are many others as well.

The way to answer a problem like this is to look for a pattern.

You know that if you divide the number by 2, there will be a remainder of 1. This eliminates half of ALL numbers!!! 2 goes evenly into all *even* numbers, so your number has to be *odd. *The number also has to be *bigger than 2*.

So, start with the list of all odd numbers greater than 2:

3,5,7,9,11,13,15,17,19,21 . . . .

Next, you know that when you divide the number by 3, there is a remainder of 2. So, any number that is a *multiple of 3* (it's in the 3 times table) can't work because there would be no remainder.

get rid of all the multiples of 3 -- I'll put them in bold

**3**,5,7,**9**,11,13,**15**,17,19,**21** . . . .

Now test the others to see if you can find a pattern:

Test 5 by dividing it by 3 -- 1 r2 **yes**

7? 2 r1 **no**

11? 3 r2 **yes**

13? 4 r1 **no**

17? 5 r2 **yes**

19? 6 r1 **no**

You can see the pattern.

5 divided by 2 equals 2 with a remainder of 1.

5 divided by 3 equals 1 with a remainder of 2.