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ya3sobi Associate
Joined: 01 Apr 2005 Posts: 2
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Posted: Fri Apr 01, 2005 12:06 pm Post subject: Help on Free Quiz March 21 |
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Hi all,
any help with the following two questions.
Thanks in Advance
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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ksb GMAT Tutor
Joined: 28 Dec 2004 Posts: 54 Location: Chennai, India
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Posted: Sat Apr 02, 2005 1:44 pm Post subject: answer to the second question |
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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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ya3sobi Associate
Joined: 01 Apr 2005 Posts: 2
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Posted: Mon Apr 04, 2005 7:45 am Post subject: |
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Thanks a lot.  |
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mahe Associate
Joined: 15 Mar 2005 Posts: 2
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Posted: Mon Apr 04, 2005 10:22 pm Post subject: |
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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. ' '
Intelligent Question setter (Choice setter)
I would be grateful if someone extend this thought and solve this problem!!!! |
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Dan Associate
Joined: 22 Jan 2005 Posts: 3
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Posted: Sat Apr 09, 2005 11:29 am Post subject: |
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| 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. ' '
Intelligent Question setter (Choice setter)
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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amanthin Associate
Joined: 13 Sep 2005 Posts: 2 Location: Colorado, USA
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Posted: Sat Sep 24, 2005 11:18 pm Post subject: |
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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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shaji Senior Consultant
Joined: 25 Apr 2005 Posts: 34
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Posted: Sun Sep 25, 2005 9:15 pm Post subject: |
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| 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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amanthin Associate
Joined: 13 Sep 2005 Posts: 2 Location: Colorado, USA
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Posted: Mon Sep 26, 2005 10:51 pm Post subject: |
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| 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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