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

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

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

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.

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

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

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

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?

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.

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.

Yes, that's right!

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