What would be your approach to solve this problem : GMAT Problem Solving (PS)
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# What would be your approach to solve this problem

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Manager
Status: Only GMAT!!
Joined: 17 Sep 2010
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Kudos [?]: 42 [1] , given: 24

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12 Jul 2011, 12:30
1
KUDOS
00:00

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Question Stats:

40% (03:56) correct 60% (00:21) wrong based on 5 sessions

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How many integers "k" greater than 100 and less than 1000 are there such that if the hundreds and the units digits of "k" are reversed, the resulting integer is k+99?

A) 50
B) 60
C) 70
D) 80
E) 90
[Reveal] Spoiler: OA
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Joined: 11 Sep 2010
Posts: 14
Location: India
GMAT 1: 620 Q49 V25
Followers: 0

Kudos [?]: 5 [0], given: 3

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13 Jul 2011, 07:19
vivgmat wrote:
How many integers "k" greater than 100 and less than 1000 are there such that if the hundreds and the units digits of "k" are reversed, the resulting integer is k+99?

A) 50
B) 60
C) 70
D) 80
E) 90

Numbers will be like 102 => 201 = 102 + 99
203 => 302 = 103 + 99

so the hundereth digit and units digit are consecutive where unit digit is bigger than hundred digit.
There will be eight pairs of such numbers
for every pair there will 10 numbers like for 1&2 => 102, 112,132,142,152, 162,172,182,192.

Total = 8 *10 = 80 hence D.
Manager
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Kudos [?]: 107 [2] , given: 94

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13 Jul 2011, 07:36
2
KUDOS
k=100x + 10y + z
99+k= 100z + 10y + x
thus 99+ 100x + 10y + z = 100z + 10y + x
99= 99z - 99x
so z = x+1
z must be no more than 8 , if it is 9 for example so x=10 and that impossible
so we have 8 possibility in each 100 and we have 10 hundreds in 1000
so 8*10= 80
you are welcome man
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How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground? - Jubran Khaleel Jubran

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Kudos [?]: 1708 [3] , given: 376

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13 Jul 2011, 08:01
3
KUDOS
Silver89 wrote:
k=100x + 10y + z
99+k= 100z + 10y + x
thus 99+ 100x + 10y + z = 100z + 10y + x
99= 99z - 99x
so z = x+1
z must be no more than 8 , if it is 9 for example so x=10 and that impossible
so we have 8 possibility in each 100 and we have 10 hundreds in 1000
so 8*10= 80
you are welcome man

I solved it the same way;

If k=xyz

x+1=z
The numbers will be:

1@2 : "@" can be 0,1,2,3,4,5,6,7,8,9. Count=10
2@3 : "@" can be 0,1,2,3,4,5,6,7,8,9. Count=10
3@4 : "@" can be 0,1,2,3,4,5,6,7,8,9. Count=10
4@5
5@6
6@7
7@8
8@9 : "@" can be 0,1,2,3,4,5,6,7,8,9. Count=10

10*8=80

Ans: "D"
_________________
Manager
Status: Trying to survive
Joined: 29 Jun 2011
Posts: 183
GMAt Status: Quant section
Concentration: Finance, Real Estate
Schools: WBS (D)
GMAT Date: 12-30-2011
GPA: 3.2
Followers: 11

Kudos [?]: 107 [0], given: 94

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13 Jul 2011, 08:06
hehe
nice solving approach fluke
and thanks for the kudos man
_________________

How can i lose my faith in life's fairness when i know that the dreams of those who sleep on the feathers are not more beautiful than the dreams of those who sleep on the ground? - Jubran Khaleel Jubran

Re: What would be your approach to solve this problem   [#permalink] 13 Jul 2011, 08:06
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