divisibility

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mav800rick: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Sun Apr 27, 2008 4:22 pm

by mav800rick » Thu May 29, 2008 4:29 pm

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 ?
A. 15
B. 16
C. 17
D. 18
E. 19

-Simple question....OA is (C). i am sure its B.

any thoughts?

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Source: Beat The GMAT — Problem Solving | Thu May 29, 2008 4:29 pm

egybs: Master | Next Rank: 500 Posts; Posts: 178; Joined: Wed May 14, 2008 3:51 pmFollowed by: 1 member; Thanked: 16 times; GMAT Score:99%+

by egybs » Thu May 29, 2008 4:36 pm

It takes 30 seconds to write out all the possibilities:

1,4,7,10,13,16,19,22,25,28,31,34,37,40,43,46,49

Count 'em up and you got 17. Answer is C.

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zacharyz: Senior | Next Rank: 100 Posts; Posts: 68; Joined: Thu Feb 14, 2008 4:11 pm; Thanked: 7 times

by zacharyz » Thu May 29, 2008 7:36 pm

Maybe forgot the 1?

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netigen: Legendary Member; Posts: 631; Joined: Mon Feb 18, 2008 11:57 pmFollowed by: 3 members; Thanked: 29 times

by netigen » Thu May 29, 2008 9:34 pm

No need to write the possibilities:

1. first multiple of 3 = 0
2. last multiple of 3 = 48

total multiples from 0 to 50 = 17 = our answer

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getneonow: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Fri Mar 14, 2008 1:56 am; Location: India; Thanked: 4 times; GMAT Score:730

by getneonow » Thu May 29, 2008 10:08 pm

The numbers are represented by 3k +1 where k = 0,1,2....

the max value K can take is k = 16 for the number 49

so from 0 to 16 there would be 17 numbers ..which is the reqd answer

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