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# Are you ready for GMAT? then try this

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Manager
Joined: 16 Feb 2010
Posts: 225
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03 May 2010, 15:37
I am seriously stuck with the following one:

Question (4) page 23:

How many positive integers less than 10,000 ae there in which the sum of the digits equalis 5?

A) 31
B) 51
C) 56
D) 62
E) 93

couldnt find a way to solve this one so went ahead and listed all possible options (yes there are 56 answers!!!) but any1 who can provide the proper solution please do so !
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03 May 2010, 16:01
zisis wrote:
I am seriously stuck with the following one:

Question (4) page 23:

How many positive integers less than 10,000 ae there in which the sum of the digits equalis 5?

A) 31
B) 51
C) 56
D) 62
E) 93

couldnt find a way to solve this one so went ahead and listed all possible options (yes there are 56 answers!!!) but any1 who can provide the proper solution please do so !

Discussed here: integers-less-than-85291.html
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05 May 2010, 08:04
anyone get this one? I know i'm missing something very silly here, it's just not clicking, please help me out.

If k is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT
(A) 8
(B) 9
(C) 12
(D) 16
(E) 17
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Math Expert
Joined: 02 Sep 2009
Posts: 38985
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Kudos [?]: 106431 [0], given: 11626

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05 May 2010, 08:20
firang wrote:
anyone get this one? I know i'm missing something very silly here, it's just not clicking, please help me out.

If k is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT
(A) 8
(B) 9
(C) 12
(D) 16
(E) 17

$$\frac{35^2-1}{k}=\frac{(35-1)(35+1)}{k}=\frac{34*36}{k}=\frac{2^3*3^2*17}{k}$$

From the answer choices only $$16=2^4$$ is not a factor of numerator, hence in case $$k=16$$, $$\frac{35^2-1}{k}$$ won't be an integer, hence k can not be 16.

 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

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05 May 2010, 08:23
Thanks for the quick response, stupid mistake on my part (a^2-b^2).

Also sorry if I violated any rule, but I thought we were discussing the pdf posted at the start of the thread that's why i posted the question here.
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Manager
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05 May 2010, 08:26
firang wrote:
anyone get this one? I know i'm missing something very silly here, it's just not clicking, please help me out.

If k is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT
(A) 8
(B) 9
(C) 12
(D) 16
(E) 17

Let me try to explain how I would approach a problem like this. First take a quick look about the answer. What do you notice?
you have a prime number that is 17 and the other are multiples of 4 or 3.
Then I did 35*35 - 1 = 1224
First, I find out if the number is divisible by 17. Because if not you have your answer here.
1224/ 17 = 72
cross out option e
now you have to check with of the answer is multiple of 72, that is easy
A and B are multiple
C too
so the answer must be D

Other option you can know that 1224 is multiple of 9 because the digits of the numbers is multiple of 9 and of course is multiple of 2. To be multiple of 4 the last digit must be multiple of 4 in this case 24 is multiple of 4. For eight the last 3 digits. (224/8 = 28)

for me the first way was really fast, but of course other approaches must be there
so I am open to suggestions!!
I hope it helps
Manager
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05 May 2010, 08:27
Bunuel wrote:
firang wrote:
anyone get this one? I know i'm missing something very silly here, it's just not clicking, please help me out.

If k is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT
(A) 8
(B) 9
(C) 12
(D) 16
(E) 17

$$\frac{35^2-1}{k}=\frac{(35-1)(35+1)}{k}=\frac{34*36}{k}=\frac{2^3*3^2*17}{k}$$

From the answer choices only $$16=2^4$$ is not a factor of numerator, hence in case $$k=16$$, $$\frac{35^2-1}{k}$$ won't be an integer, hence k can not be 16.

As always, nice approach to the problem Bunuel
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05 May 2010, 10:46
Bunuel wrote:
firang wrote:
anyone get this one? I know i'm missing something very silly here, it's just not clicking, please help me out.

If k is an integer, and (35^2-1)/k is an integer, then k could be each of the following, EXCEPT
(A) 8
(B) 9
(C) 12
(D) 16
(E) 17

$$\frac{35^2-1}{k}=\frac{(35-1)(35+1)}{k}=\frac{34*36}{k}=\frac{2^3*3^2*17}{k}$$

From the answer choices only $$16=2^4$$ is not a factor of numerator, hence in case $$k=16$$, $$\frac{35^2-1}{k}$$ won't be an integer, hence k can not be 16.

 ! Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/No posting of PS/DS questions is allowed in the main Math forum.

RESPECT !!!

i did the following very similar to the above but mine was longer:
35*35 - 1 = 1224 then found the prime numbers of 1224 and realised that 16 (=2^4) is not part of 1224
but clearly the best way is Bunuel's ! kudos
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09 May 2010, 15:20
thanks for the material
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14 May 2011, 22:36
great work. +1.

does anyone have something similar for the verbal section?
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23 May 2011, 16:30
Ah geez, only got about half right... I guess I'm not as ready as I thought I would
Re: Are you ready for GMAT? then try this   [#permalink] 23 May 2011, 16:30

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