An orderly and his pegs

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zaarathelab: Master | Next Rank: 500 Posts; Posts: 217; Joined: Sun Jan 10, 2010 1:39 pmFollowed by: 1 member; Thanked: 7 times

by zaarathelab » Thu Dec 15, 2011 11:57 am

An orderly has a number of pegs to peg in a row. At first he tried to peg 5 in each row, then 6, then 8 and then 12, but had always 1 left. On trying 13 he had none left. What is the smallest number of pegs that he could have had?

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Source: Beat The GMAT — Problem Solving | Thu Dec 15, 2011 11:57 am

kanwar86: Senior | Next Rank: 100 Posts; Posts: 99; Joined: Thu Jun 09, 2011 2:09 pm; Thanked: 11 times

by kanwar86 » Thu Dec 15, 2011 12:05 pm

zaarathelab wrote:An orderly has a number of pegs to peg in a row. At first he tried to peg 5 in each row, then 6, then 8 and then 12, but had always 1 left. On trying 13 he had none left. What is the smallest number of pegs that he could have had?

Is it 481?

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Kanwar

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GmatMathPro: GMAT Instructor; Posts: 349; Joined: Wed Sep 28, 2011 3:38 pmFollowed by: 54 members; Location: Austin, TX; Thanked: 236 times; GMAT Score:770

by GmatMathPro » Thu Dec 15, 2011 12:18 pm

zaarathelab wrote:An orderly has a number of pegs to peg in a row. At first he tried to peg 5 in each row, then 6, then 8 and then 12, but had always 1 left. On trying 13 he had none left. What is the smallest number of pegs that he could have had?

Let N be his number of pegs. From the first part of the problem it is clear that N is one more than a multiple of 5, 6, 8, 12. The smallest positive number that is a multiple of all of these numbers is, by definition, the least common multiple of 5, 6, 8, and 12. To find this, break the numbers down into their prime factors: 5, 2*3, 2*2*2, 2*2*3. From this we can deduce that the LCM is 2*2*2*3*5=120. Thus, the smallest non-trivial number of pegs he can have is 121. Other possible values of N can be obtained by adding 120 to this number. Mathematically, N=121+120n where n is a non-negative integer.

Now we have to find the smallest value of N such that N is divisible by 13. First, note that 121 divided by 13 is 9 remainder 4. That is, 121=13*9+4. Also, 120=13*9+3. So, every time we add 120 to 121, the remainder with respect to 13 will increase by 3: 4, 7, 10, 13. But having a remainder of 13 is the same as being divisible by 13, so we must have to add 120 three times before we get to a number that is divisible by 13. 121+3*120=

481

Pete Ackley
GMAT Math Pro
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zaarathelab: Master | Next Rank: 500 Posts; Posts: 217; Joined: Sun Jan 10, 2010 1:39 pmFollowed by: 1 member; Thanked: 7 times

by zaarathelab » Thu Dec 15, 2011 1:26 pm

oa is 481

Success = Max(Hardwork) + Min(Luck)

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