Let S be the set of all positive integers that, when divided

[Mike@Magoosh]

Let S be the set of all positive integers that, when divided by 8, have a remainder of 5. What is the 76th number in this set?
(A) 605
(B) 608
(C) 613
(D) 616
(E) 621

For a set of practice problems on sequences, as well as the OA & OE of this particular problem, see:
https://magoosh.com/gmat/2014/gmat-prac ... sequences/

Mike

[abhasjha]

To get the numbers in set S use .... N= 8A+5 . putting A=0 we get the first number as 5 .putting A = 1 we get the number = 13 ....

first of such number is 5 ...
next number 13 and next number 21 .....

An Arithmetic series then follows . (one way to look at remainder problems is all remainder problems are in a way arithmetic progessions !)

So there are 76 terms in the series with a common difference of 8.

so tn = a+(n-1)d
= 5+ (75)x8
= 5+600
= 605
Answer A

[Mathsbuddy]

If the first multiple of 8 is 0, then the 75th true multiple of 8 gives:
75 x 8 + 5 has last digit 5
Answer = A

[masoom j negi]

S = 8x + 5
So, terms in S are in A.P. with first term = 5 and common difference = 8
76th term = a + (76 - 1)d = 5+ (75)8= 605

[Scott@TargetTestPrep]

Let S be the set of all positive integers that, when divided by 8, have a remainder of 5. What is the 76th number in this set?
(A) 605
(B) 608
(C) 613
(D) 616
(E) 621

The first number in the set is 5, since 5/8 = 0 remainder 5.

The second number in the set is 5 + 1(8).

The third number in the set is 5 + 2(8).

The fourth number in the set is 5 + 3(8).

Continuing the pattern established, we see that the 76th number in the set is 5 + 75(8) = 605.

Answer: A