Difference between revisions of "2009 AMC 12A Problems/Problem 18"

Revision as of 21:27, 17 January 2019

The following problem is from both the 2009 AMC 12A #18 and 2009 AMC 10A #25, so both problems redirect to this page.

Problem

For $k > 0$ , let $I_k = 10\ldots 064$ , where there are $k$ zeros between the $1$ and the $6$ . Let $N(k)$ be the number of factors of $2$ in the prime factorization of $I_k$ . What is the maximum value of $N(k)$ ?

$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 8\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ 10$

Solution

The number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2}\cdot 2^{k+2} + 2^6$ .

For $k\in\{1,2,3\}$ we have $I_k = 2^{k+2} \left( 5^{k+2} + 2^{4-k} \right)$ . The first value in the parentheses is odd, the second one is even, hence their sum is odd and we have $N(k)=k+2\leq 5$ .

For $k>4$ we have $I_k=2^6 \left( 5^{k+2}\cdot 2^{k-4} + 1 \right)$ . For $k>4$ the value in the parentheses is odd, hence $N(k)=6$ .

This leaves the case $k=4$ . We have $I_4 = 2^6 \left( 5^6 + 1 \right)$ . The value $5^6 + 1$ is obviously even. And as $5\equiv 1 \pmod 4$ , we have $5^6 \equiv 1 \pmod 4$ , and therefore $5^6 + 1 \equiv 2 \pmod 4$ . Hence the largest power of $2$ that divides $5^6+1$ is $2^1$ , and this gives us the desired maximum of the function $N$ : $N(4) = \boxed{7}$ .

Solution 2

Notice that 2 is a prime factor of an integer $n$ if and only if $n$ is even. Therefore, given any sufficiently high positive integral value of $k$ , dividing $I_k$ by $2^6$ yields a terminal digit of zero, and dividing by 2 again leaves us with $2^7 \cdot a = I_k$ where $a$ is an odd integer. Observe then that $\boxed{\textbf{(B)}7}$ must be the maximum value for $N(k)$ because whatever value we choose for $k$ , $N(k)$ must be less than or equal to $7$ .

"Isn't this solution incomplete because we need to show that $N(k) = 7$ can be reached?"

An example of 7 being reached is 1000064. 1000064 divided by $2^7=128$ is 7813.

In fact, 1000064 is the ONLY $N(k)$ that satisfies 7. All others are 6 or lower, because if there are more zeroes, to be divisible by 128 ( $2^7$ ), the last 7 digits must be divisible by 128, but 64 isn't. Meanwhile, if there are less zeroes, we can test by division that they do not work.

Solution 3

As in the first solution, the number $I_k$ can be written as $10^{k+2} + 64 = 5^{k+2} 2^{k+2} + 2^6$ . Factor $2^6$ out of the expression to get $I_k = 2^6(1+2^{k-4}5^{k+2})$ .

You can now easily see that $I_k$ is divisible by at least 6 factors of two, from $2^6$ . Any other factors of two will come from the expression $(1+2^{k-4}5^{k+2})$ .

Make the substitution: $x=k-4$ . You now have $(1+2^{x}5^{x+6}) = (1+10^x 5^6)$

$5^6=15625$ , so the expression becomes $(1+15625\cdot10^x)$ This is valid when $x>-4$ .

Obviously, if $x$ is negative, the expression will be fractional and not contain any extra factors of two.

If $x>0$ , the value $15625\cdot10^x$ will end in a zero. When 1 is added to the expression, the expression will end in 1. Since numbers divisible by 2 end in 0,2,4,6, or 8, the expression is not divisible by 2 and will not contribute to the total.

If $x=0$ , the expression evaluates to $15626$ . Dividing out twos, you find that this value is only divisible by one factor of 2.

The six factors of two from earlier add to this factor of two to give $\textbf{(B)}\ 7\qquad$

Solution written by coolak

@@ Line 16: / Line 16: @@
 This leaves the case <math>k=4</math>. We have <math>I_4 = 2^6 \left( 5^6 + 1 \right)</math>. The value <math>5^6 + 1</math> is obviously even. And as <math>5\equiv 1 \pmod 4</math>, we have <math>5^6 \equiv 1 \pmod 4</math>, and therefore <math>5^6 + 1 \equiv 2 \pmod 4</math>. Hence the largest power of <math>2</math> that divides <math>5^6+1</math> is <math>2^1</math>, and this gives us the desired maximum of the function <math>N</math>: <math>N(4) = \boxed{7}</math>.
-== Alternate Solution ==
+== Solution 2 ==
 Notice that 2 is a prime factor of an integer <math>n</math> if and only if <math>n</math> is even. Therefore, given any sufficiently high positive integral value of <math>k</math>, dividing <math>I_k</math> by <math>2^6</math> yields a terminal digit of zero, and dividing by 2 again leaves us with <math>2^7 \cdot a = I_k</math> where <math>a</math> is an odd integer.

2009 AMC 12A (Problems • Answer Key • Resources)
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