BobsterGMAT wrote:

niks18, can you please explain the concept of what you did?

Hi

BobsterGMATThere is only one concept / formula that needs to be used here. Numbers are in AP series and the formula to find any number in an AP series is

\(T_n=a +(n-1)d\), where \(a\)=first term of the series, \(n\)=number of terms in the series and \(d\)=common difference i.e. difference between any two number in the series,

so here, numbers from 3 to 500 are in AP with \(a=3\), \(d=1\) & \(T_n=500\), we need \(n\)=number of terms. Using the formula, you get-

\(500=3+(n-1)*1 => 500=3+n-1=>n=498\)

Now in this series first number that is divisible by 7 will be 7 itself and the other number will be multiple of 7 so the new series will be 7, 14, 21,28.......\(T_n\) Hence common difference of this series \(d=7\)

to find the last number of this series divide 500 by 7, you will get 3 as remainder which means 3 is extra in 500, so removing 3 will give you a number that will be divisible by 7. ie. 500-3=497 will be divisible by 7 and will be our \(T_n\) for the series

Hence \(T_n=7+(n-1)*7 => n=71\)

Numbers that are not divisible by 7= Total Terms - Terms that are divisible by 7

\(=498-71=427\)