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M28-35

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M28-35  [#permalink]

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New post 16 Sep 2014, 00:30
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Question Stats:

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The product of a positive integer \(x\) and 377,910 is divisible by 3,300, then the least value of \(x\) is:

A. \(10\)
B. \(11\)
C. \(55\)
D. \(110\)
E. \(330\)

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Re M28-35  [#permalink]

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New post 16 Sep 2014, 00:31
Official Solution:

The product of a positive integer \(x\) and 377,910 is divisible by 3,300, then the least value of \(x\) is:

A. \(10\)
B. \(11\)
C. \(55\)
D. \(110\)
E. \(330\)


Given: \(\frac{377,910*x}{3,300}=integer\).

Factorize the divisor: \(3,300=2^2*3*5^2*11\).

Check 377,910 for divisibility by \(2^2\): 377,910 IS divisible by 2 and NOT divisible by \(2^2=4\) (since its last two digits, 10, is not divisible by 4). Thus \(x\) must have 2 as its factor (377,910 is divisible only by 2 so in order \(377,910*x\) to be divisible by \(2^2\), \(x\) must have 2 as its factor);

Check 377,910 for divisibility by 3: 3+7+7+9+1+0=27, thus 377,910 IS divisible by 3.

Check 377,910 for divisibility by \(5^2\): 377,910 IS divisible by 5 and NOT divisible by 25 (in order a number to be divisible by 25 its last two digits must be 00, 25, 50, or 75, so 377,910 is NOT divisible by 25). Thus \(x\) must have 5 as its factor.

Check 377,910 for divisibility by 11: \((7+9+0)-(3+7+1)=5\), so 377,910 is NOT divisible by 11, thus \(x\) must have 11 as its factor.

Therefore the least value of \(x\) is \(2*5*11=110\).


Answer: D
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Re: M28-35  [#permalink]

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New post 12 Jul 2016, 03:08
I think i did it with fewer steps, but maybe i was lucky? Hopefully someone can point it out.

-> 377,910/3,300
-> 37,791/330 (cancel out zeros)
-> ^ not further divisible by 10. Therefore, x must have 10.
-> 37,791/33 (divisible by 3?Yes!)
-> 12,597/11 (divisible by 11? No)
-> x must have at least 10 and 11.
-> 110
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Re: M28-35  [#permalink]

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New post 16 Sep 2017, 08:35
1
I found a slightly different way to do this.

"The product of a positive integer xx and 377,910 is divisible by 3,300, then the least value of x is:"

3300 is obviously a multiple of 10 and 100, but 377910 is only a multiple of 10. Hence we need a 10.
3300 is clearly a multiple of 11, but 33910 is not. So we need an 11.

This leaves 110 and 330 as possible choices. Since we can see that both 377910 and 3300 are multiples of 3, 110 is the answer.
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Re: M28-35  [#permalink]

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New post 26 Nov 2018, 03:11
We have to find the common factors between 377910 and 3300 till nothing is common. Whatever remains must be dividing x.
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Re: M28-35 &nbs [#permalink] 26 Nov 2018, 03:11
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