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# M23-29

Author Message
Manager
Joined: 06 Jun 2010
Posts: 161
Followers: 2

Kudos [?]: 17 [0], given: 151

M23-29 [#permalink]  01 Mar 2013, 05:55
If 1 book was left when a pile of books was stacked in rows of 5, how many books are there in all?
(1) When the books were stacked in rows of 7, 1 book was left

(2) When the books were stacked in rows of 8, 4 books were left

I solved this using Remainder approach.
From question stmt: N/5

1.N/7 Remainder is 1
so find a number such that when N/5 and N/7 gives remainder of 1. Found N=36.Sufficient

2.N/8 Remainder is 4
so N/5 gives remainder 1 and N/8 gives remainder 4.Found N =36.Sufficient.

Can someone xplain whats wrong with my approach?
Status: Tutor
Joined: 05 Apr 2011
Posts: 537
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 570 Q49 V19
GMAT 2: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Followers: 74

Kudos [?]: 394 [1] , given: 47

Re: M23-29 [#permalink]  01 Mar 2013, 08:28
1
KUDOS
shreerajp99 wrote:
If 1 book was left when a pile of books was stacked in rows of 5, how many books are there in all?
(1) When the books were stacked in rows of 7, 1 book was left

(2) When the books were stacked in rows of 8, 4 books were left

I solved this using Remainder approach.
From question stmt: N/5

1.N/7 Remainder is 1
so find a number such that when N/5 and N/7 gives remainder of 1. Found N=36.Sufficient

2.N/8 Remainder is 4
so N/5 gives remainder 1 and N/8 gives remainder 4.Found N =36.Sufficient.

Can someone xplain whats wrong with my approach?

B = 5k + 1

STAT1
B = 7t + 1
5k + 1 =7t + 1
k = 7t/5
So, Those values of t which will give you integer value are the common solution.
So, t=5,10,15,20
k = 7,14,21,28,35,42,49,56,63,...
So, B = 36, 71,...

So, Not Sufficient as there are many possible values!

STAT2
B = 8s + 4
5k + 1 = 8s + 4
k = (8s + 3)/5
Those values of s which will give you integer value of k are solution.
s = 4, 9,14,19,24,...
k = 7, 15, 23,31,39, 47, 55, 63
So, B = 36, 76,...

So, Not Sufficient as there are many possible values!

Trying for C
k = 7 and 63 are the two solutions which are possible (actually there are more...)
B = 36, 316

Hope it helps!
_________________
Re: M23-29   [#permalink] 01 Mar 2013, 08:28
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# M23-29

Moderators: WoundedTiger, Bunuel

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