What is this number? if I divide it by 2, the remainder is 1. If I divide by 3, the remainder is 2.
3 Answers
cburr | Middle School Teacher | (Level 2) Associate Educator
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.
enotechris | College Teacher | (Level 2) Senior Educator
5 divided by 2 equals 2 with a remainder of 1.
5 divided by 3 equals 1 with a remainder of 2.
revolution | College Teacher | (Level 1) Valedictorian
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.