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Which of the following is the lowest positive integer that

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Which of the following is the lowest positive integer that [#permalink] New post 04 Jul 2006, 07:51
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A
B
C
D
E

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Which of the following is the lowest positive integer that is divisible by:
2,3,4,5,6,7,8,9?

a. 15120
b. 3024
c. 2520
d. 1890
e. 1680

please explain strategy...
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 [#permalink] New post 04 Jul 2006, 08:20
C.

Find LCM by finding factors:

2,3,4,5,6,7,8,9 = 2,3,2*2,5,3*2,7,2*2*2,3*3

Keep only one of those that occur in different numbers and multiply:

2*3*2*5*7*2*3 = 2520
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 [#permalink] New post 04 Jul 2006, 08:26
'C' it is.

First step: B is out as we have 4 at the end.

Then start from the lowest one (E) it can not be devided by 9 - out
Then (D) can not be devided by 4 - out
Now (C) can be devided by any of these integers.

Integer is devided by 2 if it's even.
..... by 3 and 9 if sum of all digits is devided by 3.
..... by 4 if the last two digits are devided by 4.
..... by 6 if the integer is devided by 2 and by 3.
..... by 5 if the integer ends on zero or 5.

Correct me if I'm wrong.
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 [#permalink] New post 04 Jul 2006, 08:55
M8 wrote:
Correct me if I'm wrong.


Your procedure would work fine as long as the answers dont have multiples of the LCM. Say if C = 5040 which is 2520*2 then .. :toilet

So taking the LCM is the best approach.
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Last edited by giddi77 on 05 Jul 2006, 05:30, edited 1 time in total.
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 [#permalink] New post 05 Jul 2006, 15:25
amartin6165 wrote:
C.

Find LCM by finding factors:

2,3,4,5,6,7,8,9 = 2,3,2*2,5,3*2,7,2*2*2,3*3

Keep only one of those that occur in different numbers and multiply:

2*3*2*5*7*2*3 = 2520


Correct answer is C

OE, similar as yours amartin6165,

2=2^1
3=3^1
4=2^2
5=5^1*
6=(2^1)(3^1)
7=7^1*
8=2^3*
9=3^2*

take all the starts and multiply them and the answer is 2520
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 [#permalink] New post 06 Jul 2006, 14:47
M8 wrote:

What is the source?


It was one of the test question from Princeton Review...
which I got wrong...
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 [#permalink] New post 06 Jul 2006, 19:06
The LCM method is definitely the surest. Helps with an even larger set of divisors.
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Thanks,
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  [#permalink] 06 Jul 2006, 19:06
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