# Thread: Help with a little arithmetic problem

1. ## Help with a little arithmetic problem

Hello, this silly problem has me wondering.... help! Can someone explain please? Thank you!

When the positive integer n is devided by 3, the remainder is 2 and when n is divided by 5, the remainder is 1. What is the least possible value of n?  Reply With Quote

2. ## Re: Help with a little arithmetic problem

quickly playing with numbers gets me to say 11 and that is probably the best way to solve this.

We know that n-2 is divisible by 3 and n-1 is divisible by 5. So just look at numbers that end in 6 and 1 (n-1 is divisible by 5) and find one where n-2 is also divisible by 3. The first one you get is 11.  Reply With Quote

3. ## Re: Help with a little arithmetic problem Originally Posted by YaSvoboden quickly playing with numbers gets me to say 11 and that is probably the best way to solve this.

We know that n-2 is divisible by 3 and n-1 is divisible by 5. So just look at numbers that end in 6 and 1 (n-1 is divisible by 5) and find one where n-2 is also divisible by 3. The first one you get is 11.
That's the correct answer! Thank you!  Reply With Quote

4. ## Re: Help with a little arithmetic problem

When it comes to remainders, we have a nice rule that says:

If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

When the positive integer n is divided by 3, the remainder is 2

So, some possible values of n are: 2, 5, 8, 11, 14, 17, 20, 23, 26, 29,....etc

When n is divided by 5, the remainder is 1.
So, some possible values of n are: 1, 6, 11, 16, 21, 26, 31 .... etc

We can see that 11 is the smallest number that BOTH lists share

For more info on remainders, see our free video: GRE Integer Properties | Greenlight Test Prep

Cheers,
Brent  Reply With Quote

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