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

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

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

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

Mike@Magoosh wrote: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