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remainder

by GmatKiss » Mon Oct 17, 2011 12:49 pm
What is the remainder when [(11 - 1)! + 11! + (11 + 1)! + (11 + 2)! +...+ (11+ 9)!] is divided by 2^11?

A. 0
B. 1
C. 2
D. 3
E. 5
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by user123321 » Mon Oct 17, 2011 8:17 pm
ignore the post
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by user123321 » Mon Oct 17, 2011 8:37 pm
I am not sure this problem is proper.

anything above 13! factorial should be divisible 2^11
so leaving just 10!+11!+12!+13!

and this sum does not result the remainder posted under choices.
check below url

https://tinyurl.com/3uptwh7

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by Anurag@Gurome » Mon Oct 17, 2011 9:05 pm
GmatKiss wrote:What is the remainder when [(11 - 1)! + 11! + (11 + 1)! + (11 + 2)! +...+ (11+ 9)!] is divided by 2^11?
I guess the question should ask the remainder for 2^10 not 2^11.
For 2^11, it'll be impossible to determine the remainder algebraically.

# [(11 - 1)! + 11! + (11 + 1)! + (11 + 2)! +...+ (11+ 9)!]
= 10! + 11! + 12! + ... + 20!
= (10!)*(1 + 11 + 11*12 + 11*12*13 + ... + 11*12*...*20)
= (10!)*(12 + 11*12 + 11*12*13 + ... + 11*12*...*20)
= (10!)*(12)*(1 + 11 + 11*13 + 11*13*14 + ... + 11*13*...*20)
= [(2^8)*(3*5*7*9)]*[(2^2)*3]*(12 + 11*13 + 11*13*14 + ... + 11*13*...*20)
= (2^10)*(3*3*5*7*9)*(12 + 11*13 + 11*13*14 + ... + 11*13*...*20)
= (2^10)*(ODD)*(EVEN + ODD + EVEN + EVEN + ... + EVEN)
= (2^10)*(ODD)*(ODD)
= (2^10)*(ODD)

Hence, the given expression is completely divisible by 2^10.

The correct answer is A.
Last edited by Anurag@Gurome on Tue Oct 18, 2011 3:35 am, edited 1 time in total.
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by user123321 » Tue Oct 18, 2011 2:57 am
Anurag@Gurome wrote: = (2^10)*(ODD)*(ODD)
= (2^10)*(ODD)

Hence, R[Given expression/(2^11)] = R[(2^10)*(ODD)/(2^11)] = R[ODD/2] = 1

The correct answer is B.
I think there is still a problem with this one.
because 2^10*(odd)*(odd) is even
and when this even when divided by 2^11 should always leave a even remainder.

Please let me know if I did any mistake.

Thanks
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by Anurag@Gurome » Tue Oct 18, 2011 3:01 am
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Last edited by Anurag@Gurome on Tue Oct 18, 2011 3:39 am, edited 1 time in total.
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by user123321 » Tue Oct 18, 2011 3:13 am
I am still confused :(...

say, if we have to find remainder when 12 is divided by 8,
the ans is 4
we should not divide 4 in Nu & De, because cancelling out common factors will lead to wrong remainder which is 1 in this case.

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by neelgandham » Tue Oct 18, 2011 3:18 am
user123321 wrote:I am still confused :(...

say, if we have to find remainder when 12 is divided by 8,
the ans is 4
we should not divide 4 in Nu & De, because cancelling out common factors will lead to wrong remainder which is 1 in this case.

user123321
You got it wrong!

16 * 5 /32 => 5/2(plain simplification)and the remainder is 1 ? isn't it ?
In the same way (2^10)*(ODD)/2^11 => ODD/2 (It is plain simplification) and the remainder is 1,
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by Deepthi Subbu » Tue Oct 18, 2011 3:18 am
Anurag@Gurome wrote:
GmatKiss wrote:What is the remainder when [(11 - 1)! + 11! + (11 + 1)! + (11 + 2)! +...+ (11+ 9)!] is divided by 2^11?
# [(11 - 1)! + 11! + (11 + 1)! + (11 + 2)! +...+ (11+ 9)!]
= 10! + 11! + 12! + ... + 20!
= (10!)*(1 + 11 + 11*12 + 11*12*13 + ... + 11*12*...*20)
= (10!)*(12 + 11*12 + 11*12*13 + ... + 11*12*...*20)
= (10!)*(12)*(1 + 11 + 11*13 + 11*13*14 + ... + 11*13*...*20)
= [(2^8)*(3*5*7*9)]*[(2^2)*3]*(12 + 11*13 + 11*13*14 + ... + 11*13*...*20)
= (2^10)*(3*3*5*7*9)*(12 + 11*13 + 11*13*14 + ... + 11*13*...*20)
= (2^10)*(ODD)*(EVEN + ODD + EVEN + EVEN + ... + EVEN)
= (2^10)*(ODD)*(ODD)
= (2^10)*(ODD)

Hence, R[Given expression/(2^11)] = R[(2^10)*(ODD)/(2^11)] = R[ODD/2] = 1

The correct answer is B.
I am unable to follow the flow from line 6 starting [(11 - 1)! + 11! + (11 + 1)! + (11 + 2)! +...+ (11+ 9)!] . can you please elaborate ?
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by user123321 » Tue Oct 18, 2011 3:24 am
neelgandham wrote:
user123321 wrote:I am still confused :(...

say, if we have to find remainder when 12 is divided by 8,
the ans is 4
we should not divide 4 in Nu & De, because cancelling out common factors will lead to wrong remainder which is 1 in this case.

user123321
You got it wrong!

16 * 5 /32 => 5/2(plain simplification)and the remainder is 1 ? isn't it ?
In the same way (2^10)*(ODD)/2^11 => ODD/2 (It is plain simplification) and the remainder is 1,
remainder when 80 divided by 32 is not 1
32)80(2
72
-----
8

remainder is 8

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by neelgandham » Tue Oct 18, 2011 3:26 am
user123321 - makes sense ! Thanks for correcting me !
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by Anurag@Gurome » Tue Oct 18, 2011 3:38 am
user123321 wrote:say, if we have to find remainder when 12 is divided by 8,
the ans is 4
we should not divide 4 in Nu & De, because cancelling out common factors will lead to wrong remainder which is 1 in this case.
Thanks user123321!
That was really a silly mistake I made.
I think the question should ask the remainder for 2^10.
I've changed my original analysis too.
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