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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
K = 10^{26} + 2^{26}, and K is a multiple of
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Updated on: 13 Aug 2020, 17:25
Updated on: 13 Aug 2020, 17:25
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\(K = 10^{26} + 2^{26}\), and \(K\) is a multiple of \(2^n\) but NOT a multiple of \(2^{n+1}\). If n is a positive integer, what is the value of \(n+1\)?
A) 25
B) 26
C) 27
D) 28
E) 29
A) 25
B) 26
C) 27
D) 28
E) 29
Originally posted by GreenlightTestPrep on 12 Aug 2020, 13:41.
Last edited by Carcass on 13 Aug 2020, 17:25, edited 1 time in total.
Last edited by Carcass on 13 Aug 2020, 17:25, edited 1 time in total.
Edited by Carcass
Re: [m]K = 10^{26} + 2^{26}[/m], and [m]K[/m] is a multiple of [
[#permalink]
Updated on: 13 Aug 2020, 09:15
Updated on: 13 Aug 2020, 09:15
10^26 + 2^26
=(5^26)(2^26)+(2^26)
=(2^26)(5^26+1)
Now the question is whether 5^26+1 also has 2 as a factor?
5^1=5
5^2=25
5^3=125
...so on so notice that any 5 to some positive power ends with 5. Therefore 5^26+1 will end with a 6. That is 5^26+1 is even.
Therefore 2 is a factor of 5^26+1
Also (5^26+1)/2 will end with a 3 therefore (5^26+1)/2 is an odd number.
Therefore (5^26+1)/2 doesn't have 2 as a factor.
=(2^26)(5^26+1)
=(2^27)((5^26+1)/2)
Therefore
2^n = 2^27
n=27
n+1=28
Final Answer: D
=(5^26)(2^26)+(2^26)
=(2^26)(5^26+1)
Now the question is whether 5^26+1 also has 2 as a factor?
5^1=5
5^2=25
5^3=125
...so on so notice that any 5 to some positive power ends with 5. Therefore 5^26+1 will end with a 6. That is 5^26+1 is even.
Therefore 2 is a factor of 5^26+1
Also (5^26+1)/2 will end with a 3 therefore (5^26+1)/2 is an odd number.
Therefore (5^26+1)/2 doesn't have 2 as a factor.
=(2^26)(5^26+1)
=(2^27)((5^26+1)/2)
Therefore
2^n = 2^27
n=27
n+1=28
Final Answer: D
Originally posted by chacinluis on 12 Aug 2020, 17:41.
Last edited by chacinluis on 13 Aug 2020, 09:15, edited 1 time in total.
Last edited by chacinluis on 13 Aug 2020, 09:15, edited 1 time in total.
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: [m]K = 10^{26} + 2^{26}[/m], and [m]K[/m] is a multiple of [
[#permalink]
13 Aug 2020, 09:09
13 Aug 2020, 09:09
1
GreenlightTestPrep wrote:
\(K = 10^{26} + 2^{26}\), and \(K\) is a multiple of \(2^n\) but NOT a multiple of \(2^{n+1}\). If n is a positive integer, what is the value of \(n+1\)?
A) 25
B) 26
C) 27
D) 28
E) 29
A) 25
B) 26
C) 27
D) 28
E) 29
IMPORTANT: later in my solution, we need to recognize that 5^b will end in 25 for all integer values of b greater than 0.
For example, 5^2 = 25
5^3 = 125
5^4 = 625
5^5 = 3125
5^6 = XXX25 etc....
So......
K = 10^26 + 2^26
= 2^26(5^26 + 1)
= 2^26(XXXX25 + 1) [aside: XXXX25 denotes some number ending in 25]
= 2^26(XXXX26)
= 2^26[2(XXXX3)] [Since XXX26 is EVEN, we can factor out a 2]
= 2^27[XXXX3]
Since XXXX3 is an ODD number, we can't factor out any more 2's.
This means K is a multiple of 2^27, but K is NOT a multiple 2^28.
In other words, n = 27
What is the value of n + 1?
Since n = 27, we can conclude that n + 1 = 27 + 1 = 28
Answer: D
Cheers,
Brent
