The function f is defined for all non-negative integers x by the follo

Question

The function f is defined for all non-negative integers x by the following rule: f(x) = 10...064, where x represents the number of zeros between the "1" and the "64". For example, f(0) = 164, f(1) = 1064, etc. If f(a) = 2^n*k, where k is a positive odd integer and n is a positive integer, what is the maximum possible value of n?

A. 5
B. 6
C. 7
D. 8
E. 9

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Official Answer

C

A. 5
B. 6
C. 7
D. 8
E. 9

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Pankaj0901 · Answer

I am a bit confused with the question. Is 2^n*k be read as 2^(n*k) or (2^n)*k. Sorry if this is a lame question. But is there any rule (like left to right etc. I know BODMAS, but not sure how are "exponents" given priority!) to identify these kinds of terms?

FauleKatze · Answer

f(x) = 64+ 10^(x+2) = 2^6 +  
Let's assume x=6;
2^6 
Now we know that 5 to the power of anything will be an odd integer, and that 2^2 is even. Even * Odd = Even.
Even + 1 = Odd. Therefore:   = odd, and can be equal to k
=> 2^n*k = (2^6)* 
=> n= 6.
 Answer: B

chetan2u · Answer

Although  FauleKatze  has made a brilliant effort in simplifying the question, the solution goes wrong towards the end.

F(a)= 2^n*k , where k is odd.
We also know that F(a) = 10...(a times 0)...064= 100...(a+2) times 0s + 64  =  10^{a+2}+64=2^{a+2}*5^{a+2}+2^6 
Now,  2^{a+2}*5^{a+2}+2^6=2^n*k , so  n=a+2

We have to ensure that the portion after taking out 2^x is an odd integer.

1) If we take out  2^{a+2}  =>  2^{a+2}(2^{6-a-2}+5^{a+2} 
 We have to maximize a , and  2^{6-a-2}  should be least but even, so  6-a-2=1......a=3 , and  n=3+2=5

2) If we take out  2^{6}  =>  2^{6}(1+2^{a+2-6}*5^{a+2})=2^n*k 
We have to check whether we can get another 2 from the term  (1+2^{a+2-6}*5^{a+2})  to increase the value of n.
Now for  1+2^{a+2-6}*5^{a+2}  to be even,  2^{a+2-6}=1.....a=4 .
 2^{6}(1+2^{a+2-6}*5^{a+2})=2^n*k 
 2^{6}(1+5^{4+2})=2^n*k 
 Now when we add 1 to any power of 5, we get an even number that is not divisible by 5 =>  1+5^6=2y , where y is odd. 
 REASON  :  5^6=(4+1)^6 ...When you divide this by 4 we will always get a remainder 1, so we will have to add another 4-1 or 3 to get a multiple of 4. That is,  5^x+3  is a multiple of 4, while  5^x+1  is not.
Thus,  2^{6}(1+5^{4+2})=2^6*2k=2^7*k=2^n*k 
So the maximum value of n is 7.

C

chetan2u · Answer

You cannot assume x =6, because n is always x+2, so you will always get your answer as 8 if you take x as 6

FauleKatze · Answer

chetan2u  Can you please explain how you are getting n= a+2? Because n=a+2, only if you take 2^(a+2) out as common, and not if you take 2^6 out as common.

chetan2u · Answer

f(a)=1...a times 0..64=2^n*k=( 100...a+2 times 0s )+64. Now n is nothing but the number of 0s in 10....064-64, which is  a+2

Kinshook · Answer

Given: The function f is defined for all non-negative integers x by the following rule: f(x) = 10...064, where x represents the number of zeros between the "1" and the "64". For example, f(0) = 164, f(1) = 1064, etc. 
Asked: If f(a) = 2^n*k, where k is a positive odd integer and n is a positive integer, what is the maximum possible value of n?

f(x) = 10^{x+2} + 64 = 2^{x+2}*5^{x+2} + 2^6  
 f(a) = 2^{a+2}*5^{a+2} + 2^6 = 2^6 (2^{a-4}*5^{a+2} + 1) = 2^n *k

For a=4;  f(4) = 2^6(5^6+1) = 2^6(15625+1) = 2^6 * 15626 = 2^7 * 7813 
For all other values of a;  2^{a-4}*5^{a+2} + 1 is  odd and is not a multiple of 2.

Maximum value of n = 7

IMO C

abhinavsodha800 · Answer

Hi Chentan,

I didnot got this line  We have to maximize a, but it has to be less than 6  . Why a needs to be less than 6

mxgrlch · Answer

abhinavsodha800  I also don't understand this. Did you figure it out?

chetan , could you help here?
Why does a have to be less than 6?

chetan2u · Answer

Hi  abhinavsodha800  and  mxgrlch

I have changed the solution a bit, so please go through it. You will find it a bit easier to understand. Do let me know if you have any query.

abhinavsodha800 · Answer

Hi Chetan , thanks for the explaination. so as per you updated post we need to take the even term out so that power of 2 gets increased . if value of a will be greater than 6 then whole term will be odd. so we are keeping "a" equal to 4 so that odd + odd becomes even . let me know my understanding is correct and 1 + 5 will always be in the form of 2 * odd term.

sahuayush · Answer

Yes, that's right!

Posted from my mobile device

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The function f is defined for all non-negative integers x by the follo

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