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# If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the

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If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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03 Dec 2013, 16:28
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If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the what is the remainder when x is divided by 17?
A. 0
B. 4
C. 7
D. 10
E. 14

Please explain, I thought x = 2900! - 2890!; so when you can get a multiple of 17 then x could be evenly divided? Please correct my concept. Thanks.
[Reveal] Spoiler: OA

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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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03 Dec 2013, 18:51
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smartyman wrote:
If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the what is the remainder when x is divided by 17?
A. 0
B. 4
C. 7
D. 10
E. 14

Please explain, I thought x = 2900! - 2890!; so when you can get a multiple of 17 then x could be evenly divided? Please correct my concept. Thanks.

x is not equal to 2900!-2890! as you have designated. But you were correct that 2890 is a multiple of 17. As 17 is a prime, the next multiple will be 2890+17 = 2907.

Thus, we see that x can be written as : (2890+1)*(2890+2)*...(2890+10)

thus, remainder when divided by 17 : 1*2*3*4*...10 = 120*6*7*8*9*10 = (120)*(42)*(720)

Now, remainder when divided by 17 : (119+1)*(34+8)*(714+6) = 1*8*6 = 48 and remainder of 48 when divided by 17 : 34+14

Thus, E.
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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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03 Dec 2013, 21:14
Please further elaborate on this section. Thanks.

I cannot quite comprehend.

thus, remainder when divided by 17 : 1*2*3*4*...10 = 120*6*7*8*9*10 = (120)*(42)*(720)

Now, remainder when divided by 17 : (119+1)*(34+8)*(714+6) = 1*8*6 = 48 and remainder of 48 when divided by 17 : 34+14

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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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03 Dec 2013, 21:50
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smartyman wrote:
Please further elaborate on this section. Thanks.

I cannot quite comprehend.

thus, remainder when divided by 17 : 1*2*3*4*...10 = 120*6*7*8*9*10 = (120)*(42)*(720)

Now, remainder when divided by 17 : (119+1)*(34+8)*(714+6) = 1*8*6 = 48 and remainder of 48 when divided by 17 : 34+14

Imagine we have to find the remainder when 12*13 is divided by 7. Let's write 12*13 = (7+5)*(7+6) = $$7^2+7*6+7*5+5*6$$. As each term except the last one is a multiple of 7, hence, the remainder when 12*13 is divided by 7 is the same as when 5*6 is divided by 7

Thus, $$Remainder(\frac{12*13}{7}) = Remainder(\frac{5*6}{7}) \to Remainder(\frac{30}{7}) = 2$$

Note that 30 can be written as $$7*4+2$$.

This same logic can be extended to a chain of products:

For example, find remainder when $$11*12*13*15*16*17$$ is divided by 7

By the above logic :$$Remainder(\frac{11*12*13*15*16*17}{7}) = Remainder(\frac{(7+4)*(7+5)*(7+6)*(7*2+1)*(7*2+2)*(7*2+3)}{7}) --> Remainder(\frac{4*5*6*1*2*3}{7})$$ ..$$Remainder(\frac{20*36}{7}) = Remainder(\frac{6*1}{7}) = 6.$$

Note that $$20 = 7*2+6 ; 36 = 7*5+1$$

Let me know if any other doubt still persists.
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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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14 Dec 2013, 03:16
Just curious, is this truly 600-700 level or 700+?
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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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01 Oct 2016, 02:29
Hello from the GMAT Club BumpBot!

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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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01 Oct 2016, 04:21
Awesome question. Could not figure it out, but great explanation above

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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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27 Nov 2016, 23:53
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smartyman wrote:
If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the what is the remainder when x is divided by 17?
A. 0
B. 4
C. 7
D. 10
E. 14

$$\frac{2891*2892*2893*...*2899*2900}{17} = \frac{1*2*3*5*6*7*8*9*10}{17} = \frac{(1*2*3*6)*(5*4)*(8*9)*(7*10)}{17} = \frac{36*20*72*70}{17} = \frac{2*3*4*2}{17} = \frac{48}{17}$$
= -$$3 (mod 17)$$ = $$14 (mod 17)$$

Remainder is 14.

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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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28 Nov 2016, 07:20
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smartyman wrote:
If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the what is the remainder when x is divided by 17?
A. 0
B. 4
C. 7
D. 10
E. 14

Please explain, I thought x = 2900! - 2890!; so when you can get a multiple of 17 then x could be evenly divided? Please correct my concept. Thanks.

$$\frac{2891}{17}$$ = Remainder 1
$$\frac{2892}{17}$$ = Remainder 2
$$\frac{2893}{17}$$ = Remainder 3
$$\frac{2894}{17}$$ = Remainder 4
$$\frac{2895}{17}$$ = Remainder 5
$$\frac{2896}{17}$$ = Remainder 6
$$\frac{2897}{17}$$ = Remainder 7
$$\frac{2898}{17}$$ = Remainder 8
$$\frac{2899}{17}$$ = Remainder 9
$$\frac{2900}{17}$$ = Remainder 10

$$\frac{1*2*3*4*5*7}{17} = \frac{5040}{17}$$ = Remainder 8
$$\frac{8*9*10}{17} = \frac{720}{17}$$ = Remainder 6

Again, $$\frac{8*6}{17}$$ = Remainder 14

So, Answer will be (E) 14
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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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13 Aug 2017, 11:01
Using Remainder Th.
Since 2890 = 289*10, and 289 is square of 17. Thus 2890 is perfectly divisible by 17.
Now
2891 will leave 1 as remainder.
2892 will leave 2 as remainder.
2893 will leave 3 as remainder.
2894 will leave 4 as remainder.
2895 will leave 5 as remainder.
2896 will leave 6 as remainder.
2897 will leave 7 as remainder.
2898 will leave 8 as remainder.
2899 will leave 9 as remainder.
2990 will leave 10 as remainder.
Now$$\frac{1*2*3*4*5*6*7*8*9*10}{17}$$
Using remainder th. we can used smaller multiplication like
$$\frac{9*2}{17}$$ gives 1 as remainder,
$$\frac{6*3}{17}$$gives 1 as remainder,
$$\frac{5*7}{17}$$gives 1 as remainder,
$$\frac{4*10}{17}$$gives 6 as remainder,
and 8 is remaining,
thus ultimate remainder is
1*1*1*6*8 = 48 which when divided by 17 gives 14 as remainder.
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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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13 Aug 2017, 14:48
how to find the remainder of this division: 10! / 17?
well, take 1x9 and 2x8 and 3x7 and 4x6 and 5x10
it should be 24x21x16x9x50
then take the remainder again, => 7x4x(-1)x9x(-1)
snd again => 14

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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the   [#permalink] 13 Aug 2017, 14:48
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