Divisibility

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mitaliisrani: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Sun Mar 21, 2010 11:06 am

by mitaliisrani » Mon May 24, 2010 9:26 pm

4^41 + 4^42 + 4^43 + 4^44 + 4^45 is divisible by
A) 11
B) 31
C) 41
D) 11 and 31
E) 11& 41

OAD

I guess it has something to do with sum of all natural numbers is n(n+1)/2 but cannot figure a way...Plz help!

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Source: Beat The GMAT — Problem Solving | Mon May 24, 2010 9:26 pm

Patrick_GMATFix: GMAT Instructor; Posts: 1052; Joined: Fri May 21, 2010 1:30 amFollowed by: 98 members; Thanked: 335 times

by Patrick_GMATFix » Mon May 24, 2010 9:49 pm

To solve, you need to factor out 4^41. What's left inside is small enough that it can be calculated in under 90 seconds:

4^41 (1 + 4 + 4^2 + 4^3 + 4^4)

4^41 (1 + 4 + 16 + 64 + 256)

4^41 (341)

4^41 (11) (31) *

This number is divisible by 11 and 31 since these appear among its factors. This is why the correct answer is D

*Note that whenever in a 3-digit number, the middle digit is the sum of the 1st and 3rd, that 3-digit number is a multiple of 11. Examples include 341, 253, 594...

If you have trouble with pattern and exponent questions, consider using the Solutions Engine to generate customized drills from topic="exponents AND sequences" and difficulty="600-700 AND 700+".

Best of luck,
-Patrick

Last edited by Patrick_GMATFix on Mon May 24, 2010 9:50 pm, edited 1 time in total.

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liferocks: Legendary Member; Posts: 576; Joined: Sat Mar 13, 2010 8:31 pmFollowed by: 1 member; Thanked: 97 times

by liferocks » Mon May 24, 2010 9:50 pm

I solved it like this

4^1=4...........4%11=4
4^2=16........16%11=5
4^3=64........64%11=9
4^4=256.....256%11=3
4^5=1024..1024%11=1

So every power which is multiple of 5 gives 1 as reminder
hence
(4^41 + 4^42 + 4^43 + 4^44 + 4^45)%11=(4+4^2+4^3+4^4+4^5)%11=(4+5+9+3+1)%11=0

4^1=4...........4%31=4
4^2=16........16%31=16
4^3=64........64%31=2
4^4=256.....256%31=8
4^5=1024..1024%31=1

So every power which is multiple of 5 gives 1 as reminder
hence
(4^41 + 4^42 + 4^43 + 4^44 + 4^45)%31=(4+4^2+4^3+4^4+4^5)%31=(4+16+2+8+1)%31=0

hence 4^41 + 4^42 + 4^43 + 4^44 + 4^45 is divisible by both 11 and 31..didnt tried with 41 after that

Ans option D

This took me more than 2 minutes.will wait for other guys to pitch in for a better solution.

"If you don't know where you are going, any road will get you there."
Lewis Carroll

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members; Thanked: 135 times

by selango » Mon May 24, 2010 10:02 pm

The best approach will be like,

4^41(1+4+4^2+4^3+4^4)

=4^41(341)

=4^41*11*31

4^41 will be the multiple of 4.But there is no multiple of 4 in options.

This leaves behind only 11 and 31

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mitaliisrani: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Sun Mar 21, 2010 11:06 am

by mitaliisrani » Wed May 26, 2010 4:04 am

Thank you fr the explanations..much appreciated!

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mitaliisrani: Senior | Next Rank: 100 Posts; Posts: 56; Joined: Sun Mar 21, 2010 11:06 am

by mitaliisrani » Wed May 26, 2010 4:25 am

Patrick_GMATFix wrote:To solve, you need to factor out 4^41. What's left inside is small enough that it can be calculated in under 90 seconds:

4^41 (1 + 4 + 4^2 + 4^3 + 4^4)

4^41 (1 + 4 + 16 + 64 + 256)

4^41 (341)

4^41 (11) (31) *

This number is divisible by 11 and 31 since these appear among its factors. This is why the correct answer is D

*Note that whenever in a 3-digit number, the middle digit is the sum of the 1st and 3rd, that 3-digit number is a multiple of 11. Examples include 341, 253, 594...

If you have trouble with pattern and exponent questions, consider using the Solutions Engine to generate customized drills from topic="exponents AND sequences" and difficulty="600-700 AND 700+".

Best of luck,
-Patrick

Thanks went through the Gmat Fix website...It was really helpful!

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Patrick_GMATFix: GMAT Instructor; Posts: 1052; Joined: Fri May 21, 2010 1:30 amFollowed by: 98 members; Thanked: 335 times

by Patrick_GMATFix » Wed May 26, 2010 6:45 am

mitaliisrani wrote:
Patrick_GMATFix wrote:If you have trouble with pattern and exponent questions, consider using the Solutions Engine to generate customized drills from topic="exponents AND sequences" and difficulty="600-700 AND 700+".

Best of luck,
-Patrick
Thanks went through the Gmat Fix website...It was really helpful!

Thanks for the feedback. It's meant to be used in conjunction with the type of community discussion you get here at BTG

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