Integer Question - Looking for values for K

[jkcm25]

Q: If "K" is an integer and 3 is the remainder when K is divided by 5, how many values for K are possible when 0<k<100?

I thought the answer was "19", and it turns out to be "20". Could someone explain to me why the smallest value for K is "3"?
I started my lowest value for K as 8 where 8 / 5 = remainder 3.

How is it possible that K's lowest value could equal to "3" when clearly 3 / 5 does not yield to an integer answer with the remainder 3? Could someone refute my obvious flawed reasoning?

[sana.noor]

its 20, start with 3 and then 8. 3 will be the smallest value and you can divide it as 0/5 = (5)(0) + 3. 3 is the remainder when divided by 5. it means in every 10 natural numbers their are 2 such numbers. less than 100 means that the last number will be 98. therefore, (2)(10) = 20

[jkcm25]

Hi sorry, this did not really answer my question why the lowest value for K = 3.
I understand the subsequent integer answers STARTING from "K" integers: 8,13,18,..... up to 98 when these "K" integers are divided 5 it yields 3 as a remainder.

How is it that K integer "3" when divided by "5" could yield a remainder of "3"? So 3 divided 5= integer, Remainder 3???

[jkcm25]

I've skimmed your answer so fast. I missed your point. Thanks!

[sana.noor]

5 multiply by 0 will lead to answer 0...Remember: 0 is the multiple of every number thus u can divide 3 with 5. 3/5 = (5)(0) + 3...so include 3 and the right answer is 20

[jkcm25]

Thanks for the clarification. I got it

[Brent@GMATPrepNow]

Q: If "K" is an integer and 3 is the remainder when K is divided by 5, how many values for K are possible when 0<k<100?

I thought the answer was "19", and it turns out to be "20". Could someone explain to me why the smallest value for K is "3"?
I started my lowest value for K as 8 where 8 / 5 = remainder 3.

How is it possible that K's lowest value could equal to "3" when clearly 3 / 5 does not yield to an integer answer with the remainder 3? Could someone refute my obvious flawed reasoning?

Most students are good a listing some possible values that satisfy given information regarding remainders. However, they often miss the smallest possible value (in this case the value of 3)

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.

Cheers,
Brent