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# What is the lowest possible integer that is divisible by

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Joined: 30 Jun 2007
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What is the lowest possible integer that is divisible by [#permalink]

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10 Jul 2007, 16:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the lowest possible integer that is divisible by each of the integers 1 through 7, inclusive?
a. 420
b. 840
c. 1260
d. 2520
e.5040

I know I can solve it the long way, but how do I solve it if something like this comes up on the test? Thanks,

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Director
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10 Jul 2007, 16:06
What do you consider the long way?

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Director
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10 Jul 2007, 16:12
They are all divisible by 1, 2, and 5 for obvious reasons. That leaves 3, 4, 6 and 7.

3-They are all divisible by three since the sum of the digits equal three.
4-They are all divisible by four since the last two digits of each are divisible by four
6-when a number is divisible by 3 and 2 it is divisible by 6.
7-I don't have a rule but this narrows it to 420.

I did this in my head in 30 seconds. Not bragging, but just showing that if you know these rules it makes these problems much easier.

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10 Jul 2007, 16:36
Right, the answer is 420. Thanks, I'll be emphasizing more on those rules!

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10 Jul 2007, 19:17
Use the laws of divisbility
1: -> a to e are divisible by 1
2: -> a to e are divisible by 2
3: -> a to e are divisble by 3
4: -> a to e are divisible by 4
5: -> a to e are divisible by 5
6: -> a to e are divisible by 6
7: -> a to e are divisible by 7

So the lowest possible integer = 420 = A

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Senior Manager
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10 Jul 2007, 21:51
I just wrote out the numbers quickly and multiplied as necessary.

1 2 3 4/2 5 6/2.3 7

6.2.5.7

60.7

420

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Director
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11 Jul 2007, 13:21
Get all the prime first:

2 * 3 * 5 * 7. Well 4 still doesn't divide.

so 2 * 2 * 3 * 5 * 7 = 420
A.

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11 Jul 2007, 14:57
asaf wrote:
Get all the prime first:

2 * 3 * 5 * 7. Well 4 still doesn't divide.

so 2 * 2 * 3 * 5 * 7 = 420
A.

the question is asking for the LCM (LCD if they were fractions).

to find the LCM, break the two numbers into primes. then find the product of all the primes factors of both numbers, using the higher power of any repeated factors

1 2 3 4 5 6 7=

1 2 3 2^2 5 2*3 7

now take the product of all the primes (using the highest power of any repeated elements)

1*2*2*3*5*7=420

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11 Jul 2007, 14:58
ywilfred wrote:
Use the laws of divisbility
1: -> a to e are divisible by 1
2: -> a to e are divisible by 2
3: -> a to e are divisble by 3
4: -> a to e are divisible by 4
5: -> a to e are divisible by 5
6: -> a to e are divisible by 6
7: -> a to e are divisible by 7

So the lowest possible integer = 420 = A

i think that might be 'the long way' joselord was referring to

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11 Jul 2007, 19:52
it is 420 using simple LCM calculation.

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12 Jul 2007, 02:35
Calculate LCM, which is 420.

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12 Jul 2007, 02:35
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