Let n, k, and y be positive integers such that y n. If k*2^n = 2^y +

Question

Let n, k, and y be positive integers such that  y > n . If  k*2^n = 2^y + 2^n , then k could equal all of the following except:

A. 33
B. 65
C. 123
D. 257
E. 513

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CareerGeek · Answer

k*2^n = 2^y + 2^n 
-->  k*2^n = 2^n(1 + 2^{y - n}) 
-->  k = 1 + 2^{y - n} 
So, possible value of k is 1 more than any positive power of 2 (Since  y > n )

A. 33 -->  1 + 2^5  -->  Possible 
B. 65 -->  1 + 2^6  -->  Possible 
C. 123 -->  -5 + 2^7  -->  Not Possible 
D. 257 -->  1 + 2^8  -->  Possible 
E. 513 -->  1 + 2^9  -->  Possible

IMO Option C

freedom128 · Answer

Dividing both sides of  k*2^n = 2^y + 2^n  with  2^n , one obtains

k = 1 + 2^{y-n}  , where  k, n, y, (y-n)   are positive integers

A. 33  = 2^5+1 
B. 65  = 2^6+1 
 C. 123  
eq{2^7+1}  (k cannot be 123. This is the answer) 
D. 257  = 2^8+1 
E. 513  = 2^9+1

Final answer is (C)

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chetan2u · Answer

k*2^n = 2^y + 2^n.....k*2^n-2^n=2^y.....2^n(k-1)=2^y 
This will be possible only when k-1 is in the form  2^x , so k should be 1 LESS than some power of 2..
A. 33... 2^5-1 
B. 65.. .2^6-1 
C. 123... 2^7-5 ..This is NOT in the form  2^x-1 
D. 257.... 2^8-1 
E. 513... 2^9-1

C

KarishmaB · Answer

We know that exponents of 2 are involved. 
 k*2^n = 2^y + 2^n

If you feel lost with the equation, look at the options. Note that each option is very close to a power of 2.
33 is 1 more than 2^5, 65 is 1 more than 2^6. 123 is 5 less than 2^7. 257 is 1 more than 2^8 and 513 is 1 more than 2^9.
By this pattern, the only odd one out is 123 so that is most likely the answer.

To solve it, try to plug in 33 to see if that works and if it does, how does it work. 
 33*2^n = 2^y + 2^n 
 (2^5 + 1)*2^n = 2^y + 2^n 
 2^7 + 2^n = 2^y + 2^n  - this works

We see that 33 is 1 more than 2^5 and that is why we continue to get a 2^n on LHS too and we get another power of 2 on LHS. Options B, D and E are also 1 more than a power of 2 so they will also work.

Only option (C) will not work.

This  video discusses exponents .

Let n, k, and y be positive integers such that y > n. If k*2^n = 2^y +

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