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 Post subject: Help on Free Quiz March 21
PostPosted: Fri Apr 01, 2005 6:36 am 
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Joined: Fri Apr 01, 2005 6:30 am
Posts: 2
Hi all,

any help with the following two questions.

Thanks in Advance :wink:

3 The sum and product of three consecutive terms of an A.P. are 27 and 504 respectively. Which of the following could be the common difference of the terms of the A.P?
-3
3
-5
-4
6

5 Which of the following numbers will definitely divide a six digit positive integer xyxyxy, where x takes values between 1 and 9 and y takes values between 0 and 9?
39
28
35
41
93


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 Post subject: answer to the second question
PostPosted: Sat Apr 02, 2005 8:14 am 
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GMAT Tutor

Joined: Tue Dec 28, 2004 2:34 am
Posts: 81
Location: Chennai, India
Any number of the form xyxyxy can be written as xy(10000) + xy(100) + xy

i.e, xy(10101)

Therefore, this number xyxyxy will be divisible by all divisors of 10101.

10101 = 111 * 91 = 37 * 3 * 13 * 7

Therefore, any combination of products of 2 or more of 37, 3, 13 and 7 will divide this number.

3 * 13 = 39.

Hence, choice A.

_________________
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 Post subject:
PostPosted: Mon Apr 04, 2005 2:15 am 
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Joined: Fri Apr 01, 2005 6:30 am
Posts: 2
Thanks a lot. :)


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 Post subject:
PostPosted: Mon Apr 04, 2005 4:52 pm 
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Joined: Tue Mar 15, 2005 12:40 pm
Posts: 2
Solution for Q5 by KSB was a good one.

Heres My approach though it fail to get right answer =>>>>

Anyone Remember Divisibility by 3 rule:
If the sum of the digits is divisible by three, the number is divisible by 3
add all the digits in xyxyxy we get 3(x+y).This is divisible by 3.

One can conclude that the divisor could also divisible by 3.
Now look at the choices only Options 1 and 5 are divisible.
So i eliminate other options.
39 = 3*13
93 = 3 * 31
Now its difficult which one to choose. ':roll:'

Intelligent Question setter (Choice setter) 8)

I would be grateful if someone extend this thought and solve this problem!!!!


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 Post subject:
PostPosted: Sat Apr 09, 2005 5:59 am 
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Joined: Sat Jan 22, 2005 8:45 am
Posts: 3
mahe wrote:
Solution for Q5 by KSB was a good one.

Heres My approach though it fail to get right answer =>>>>

Anyone Remember Divisibility by 3 rule:
If the sum of the digits is divisible by three, the number is divisible by 3
add all the digits in xyxyxy we get 3(x+y).This is divisible by 3.

One can conclude that the divisor could also divisible by 3.
Now look at the choices only Options 1 and 5 are divisible.
So i eliminate other options.
39 = 3*13
93 = 3 * 31
Now its difficult which one to choose. ':roll:'

Intelligent Question setter (Choice setter) 8)

I would be grateful if someone extend this thought and solve this problem!!!!



check extremes (minimum and maximum values). Take 101010 (x = 1 and y = 0 lowest they can take).

now do divisibility checks for 101010
by 13: delete last digit and subtract 9 times deleted; you reach 91 divisible by 13.
by 31: delete last digit and subtract 3 times deleted; you reach 83 NOT divisible by 31.

So it's 39.


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 Post subject:
PostPosted: Sat Sep 24, 2005 5:48 pm 
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Joined: Mon Sep 12, 2005 7:06 pm
Posts: 2
Location: Colorado, USA
Here is the answer to your first question...

Let the numbers be...
x x+d, x+2d
Staement 1: 3x+3d = 3(x+d) = 27
x+ d = 9
Statement 2:
x(x+d)(x+2d) = 504
x(9)(9+d) = 504 ... Substituting statement 1.
x(9+d) = 56
56 can be written as
14 * 4
28 * 2
Which would leave 9+d = 4 or 9+d = 2 ....
Therefore d could be -5 or -7...
So the answer is -5


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 Post subject:
PostPosted: Sun Sep 25, 2005 3:45 pm 
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Senior Consultant

Joined: Mon Apr 25, 2005 6:36 am
Posts: 34
amanthin wrote:
Here is the answer to your first question...

Let the numbers be...
x x+d, x+2d
Staement 1: 3x+3d = 3(x+d) = 27
x+ d = 9
Statement 2:
x(x+d)(x+2d) = 504
x(9)(9+d) = 504 ... Substituting statement 1.
x(9+d) = 56
56 can be written as
14 * 4
28 * 2
Which would leave 9+d = 4 or 9+d = 2 ....
Therefore d could be -5 or -7...

Hi;

-7 is incorrect, since x+d=9, then x can't be 28.

d=5 or -5 and the numbers are:
4,9 & 14 Or 14,9 & 4

Shaji
So the answer is -5


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 Post subject:
PostPosted: Mon Sep 26, 2005 5:21 pm 
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Joined: Mon Sep 12, 2005 7:06 pm
Posts: 2
Location: Colorado, USA
Thanks for pointing that out Shaji. I should have made an explicit statement that you need to check the conditions before picking the answers. I was just showing how you could have multiple answers...


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