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

Revision as of 19:14, 14 September 2022

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 1

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

Solution 4

Similar to the other solutions, notice that $I_k$ can be written as $10^{k+2}+64 \Rightarrow 2^{k+2}5^{k+2}+2^6$ . Factoring out $2^6$ we see that

$I_k = 2^6(2^{k-4}5^{k+2}+1)$

Notice that for $k < 4$ , $2^{k-4}$ will not be an integer, and will "steal" some $2$ 's from the $2^6$ . We don't want this to happen, since we want to maximize the exponent of $2$ . We start by considering $k = 4$ . Then

$I_k = 2^6(*5^6+1) \Rightarrow 2^6(5^6+1)$

$5^6$ is an odd number; more specifically, it ends in $25$ (all powers of $5$ after $5^1$ end in $25$ ). Therefore the value in the parentheses will be some large number that ends in $26$ . Considering the rules of divisibility, we find that $5^6+1$ is even, so it is divisible by $2$ . Now our exponent of $2$ is at $7$ . But the divisibility rule for $4$ is the last $2$ digits of the number must be divisible by $4$ . We know the last digits: $26$ , which is not divisible by $4$ . Therefore $5^6 + 1$ is divisible by $2$ , but not $4$ . Testing more values of $k$ , we find that for $k \ge 5$ , the last digit becomes $1$ , which means it is not even divisible by $2$ . Therefore the highest possible exponent of $2$ that we can reach is $7 \Rightarrow \boxed{\text{B}}$ .

Solution 5 (LTE)

$v_2(10^k+2^6)=6$ if $k>6$ and $v_2(10^k+2^6)=k$ if $k<6$ . However, if $k=6$ , then $v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)$ . By LTE, $v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3$ . Since $v_2(5^6-1)=3$ , $v_2(5^6+1)$ must equal $1$ . So, the answer is $6+1=7 \Rightarrow \boxed{\text{B}}$

@@ Line 62: / Line 62: @@
 ==Solution 5 (LTE)==
 <math>v_2(10^k+2^6)=6</math> if <math>k>6</math> and <math>v_2(10^k+2^6)=k</math> if <math>k<6</math>.
-However, if <math>k=6</math>, then <math>v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)</math>. By LTE, <math>v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3</math>. Since <math>v_2(5^6-1)=3</math>, <math>v_2(5^6+1)</math> must equal <math>1</math>. So, the answer is <math>6+1=\boxed{7}</math>
+However, if <math>k=6</math>, then <math>v_2(10^6+2^6)=v_2(2^6(5^6+1))=6+v_2(5^6+1)</math>. By LTE, <math>v_2(5^6-1)=v_2(5-1)+v_2(5+1)+v_2(6)-1=2+1+1-1=3</math>. Since <math>v_2(5^6-1)=3</math>, <math>v_2(5^6+1)</math> must equal <math>1</math>. So, the answer is <math>6+1=7 \Rightarrow \boxed{\text{B}}</math>
 == See Also ==

2009 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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