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Integers

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Source: — Problem Solving |
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by gmatinjuly » Sat May 31, 2008 5:11 pm
The anser should be 3 i.e D

Here is how I solved

To find a number that can be a multiple of all numbers from 1 to 10 we need to find the least number that can be divided by 1 to 10.

Going by this...least number would be
2 ^3 * 3 ^2 * 5 * 7 = 2520

Now there are only 3 muliplea of 2520 that are 4 digit
2520
5040
7560

Hence d
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by onlyGmat2008 » Sun Jun 01, 2008 1:19 am
Could you please explain how did you get the least number 2520?
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by jasonc » Sun Jun 01, 2008 2:17 am
1,2,3,4,5,6,7,8,9,10.
for all prime numbers, we need at least 1x of - 2,3,5,7
for the others lets break them down - 4,6,8,9,10
4=2x2
6=3x2
8=2x2x2
9=3x3
10=2x5

adding the max # of each we need we arrive at:
three 2s, two 3s, one 5, and one 7.
=2^3*3^2*5*7 = 8*9*7*5 = 2520
I beat the GMAT! 760 (Q49/V44)
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by onlyGmat2008 » Sun Jun 01, 2008 2:50 am
Thanks jasonc for the explanation.
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by molt_llest » Sun Jun 01, 2008 3:27 am
Thank you very much, now I get it!
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by AleksandrM » Sun Jun 01, 2008 3:39 pm
Very clever approach.
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