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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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New post 03 Dec 2013, 17:28
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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.
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Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the [#permalink]

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New post 03 Dec 2013, 19:51
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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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New post 03 Dec 2013, 22:14
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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New post 03 Dec 2013, 22:50
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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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New post 14 Dec 2013, 04:16
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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New post 01 Oct 2016, 05:21
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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New post 28 Nov 2016, 00:53
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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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New post 28 Nov 2016, 08:20
1
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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New post 13 Aug 2017, 12:01
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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New post 13 Aug 2017, 15:48
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
Re: If x = 2891 × 2892 × 2893 ×……………..× 2898 × 2899 × 2900, the   [#permalink] 13 Aug 2017, 15:48
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