Divisibility

[mitaliisrani]

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!

[Patrick_GMATFix]

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

[liferocks]

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.

[selango]

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

[mitaliisrani]

Thank you fr the explanations..much appreciated!

[mitaliisrani]

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!

[Patrick_GMATFix]

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 .