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m25#02

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Re: m25#02 [#permalink]

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02 Dec 2011, 00:55
There are 6 between 0 and 20 plus 1 for 10!. Answer is B.

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Re: m25#02 [#permalink]

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03 Feb 2012, 22:36
7... its important to remember that 10! is a multiple of 3
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Re: m25#02 [#permalink]

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04 Feb 2012, 09:39
duuuma wrote:
How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

As soon as you realize that 10! itself is a multiple of 3, then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 form 0 to 20, inclusive?
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.

So, (18-0)/3+1=7

Answer: B.

Hope it's clear.
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Re: m25#02 [#permalink]

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06 Mar 2012, 19:45
Why do we consider 10! here as the Q asks between 10! and 10! +20 which doesn't include end points

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Re: m25#02 [#permalink]

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06 Mar 2012, 21:54
mithun2vrs wrote:
Why do we consider 10! here as the Q asks between 10! and 10! +20 which doesn't include end points

Welcome to GMAT Club.

In fact we ARE told that 10! and 10!+20 should also be considered: "How many integers are divisible by 3 between 10! and 10!+20, INCLUSIVE?"

Hope it's clear.
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Re: m25#02 [#permalink]

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18 Sep 2012, 08:00
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This question looks a lot more complicated than it actually is. Lets break it down:
1) 10! is a multiple of 3 b/c 10x9x8x7x6x5x4x3x2x1 (ignoring the 6 and 9 for simplicity)

2) Two multiples of n added together are also a multiple of n e.g. 3+6 = 9

3) Therefore the question is really how many multiples of 3 are between 0 and 20 (of which there are 6)

4) Then the final bit of housekeeping: add one because the '0' above is actually 10!, which is a multiple of 3.

So the total is 7
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Re: m25#02 [#permalink]

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03 Dec 2012, 08:17
Bunuel wrote:
duuuma wrote:
How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

As soon as you realize that 10! itself is a multiple of 3, then the question boils down to: how many integers from some multiple of 3 to that multiple of 3 + 20, inclusive are divisible by 3?

Or: how many integers are divisible by 3 form 0 to 20, inclusive?
$$# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1$$.

So, (18-0)/3+1=7

Answer: B.

Hope it's clear.

Quote:
This rule applies to any number of consecutive integers: The product of k consecutive integers is always divisible by k factorial (K!).

between 0 and 20 we have 6 number + 1 from 10! = 7

As always, it seems difficult but is not at all
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Re: m25#02 [#permalink]

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03 Dec 2012, 10:31
duuuma wrote:
How many integers are divisible by 3 between $$10!$$ and $$10! + 20$$ inclusive?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

I'm glad to see This is the first arthematic question which I got it right on the first step in this forum,,,:)

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Re: m25#02 [#permalink]

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04 Dec 2012, 14:39
10! is for sure divisible by 3
so 10!+ 20 will have 10!+3, 10!+6,10!+9,10!+12, 10!+15 & 10!+18 total 6 + 10! which is alreday divisible by 3
so 6+1 =7

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Manager
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Re: m25#02 [#permalink]

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28 Nov 2013, 22:20
nerd wrote:
10! is for sure divisible by 3
so 10!+ 20 will have 10!+3, 10!+6,10!+9,10!+12, 10!+15 & 10!+18 total 6 + 10! which is alreday divisible by 3
so 6+1 =7

10! is divisible by 3 as 10! = 2 * 3 *4 * 5* 6* 7* 8* 9* 10
The nearest and farthest multiple in the range, 10! and 10! +18
The difference between the extremes 10! + 18 – 10! = 18
So number of multiples in the range (18/3) + 1 = 7
Hence, option (B) is the answer.

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Re: m25#02   [#permalink] 28 Nov 2013, 22:20

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