# What is this number? if I divide it by 2, the remainder is 1. If I divide by 3, the remainder is 2.

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### 3 Answers

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.

It can be anything. You know it cannot be a even number, as if it is divided by 2 , then there would be no remainder, so it must be a odd no. Also, it is not divisible by 3 like 3,6,9,12,15,18..... as it also won't have a remainder or also by 2 as it would not abide with the conditions stated above, {remainders when divisible by 2 and 3}.

The odd numbers are here: 3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37 and so on.

From this list, cancel out those that are multiples of 3.

So, 3,9,15,18,21,27,33 are all out

To test out the rest of the numbers fulfil those conditions:

5= 2R1, 1R2 (passed)

7= 3R1, 2R1 (failed)

11= 5R1, 3r2 (passed) and so. You should know by then which are the odd one out.