Difference between revisions of "2010 AMC 10A Problems/Problem 25"

Revision as of 18:22, 29 January 2011

We can find the answer by working backwards. We begin with $1-1^2=0$ on the bottom row, then the $1$ goes to the right of the equal's sign in the row above. We find the smallest value $x$ for which $x-1^2=1$ and $x>1^2$ , which is $x=2$ .

We repeat the same procedure except with $x-1^2=1$ for the next row and $x-1^2=2$ for the row after that. However, at the fourth row, we see that solving $x-1^2=3$ yields $x=4$ , in which case it would be incorrect since $1^2=1$ is not the greatest perfect square less than or equal to $x$ . So we make it a $2^2$ and solve $x-2^2=3$ . We continue on using this same method where we increase the perfect square until $x$ can be made bigger than it. When we repeat this until we have $8$ rows, we get:

$\begin{array}{ccccc}{}&{}&{}&{}&7223\\ 7223&-&84^{2}&=&167\\ 167&-&12^{2}&=&23\\ 23&-&4^{2}&=&7\\ 7&-&2^{2}&=&3\\ 3&-&1^{2}&=&2\\ 2&-&1^{2}&=&1\\ 1&-&1^{2}&=&0\\ \end{array}$

Hence the solution is the last digit of $7223$ , which is $\boxed{\textbf{(B)}\ 3}$ .

@@ Line 1: / Line 1: @@
-We can find the answer by working backwards. We begin with <math>1-1^2=0</math> on the bottom row, then the <math>1</math> goes to the right of the equal's sign in the row above. We find the smallest value <math>x</math> for which <math>x-1^2=1</math> (and that does not violate the fact that <math>1^2</math> must be the greatest perfect square less than <math>x</math>), which is <math>2</math>.
+We can find the answer by working backwards. We begin with <math>1-1^2=0</math> on the bottom row, then the <math>1</math> goes to the right of the equal's sign in the row above. We find the smallest value <math>x</math> for which <math>x-1^2=1</math> and <math>x>1^2</math>, which is <math>x=2</math>.
-We continue with the same way with the next row, but at the fourth row, we see that solving <math>x-1^2=3</math> yields <math>x=4</math>, but in that case we would have to make the <math>1^2</math> a <math>2^2</math> since <math>4 \geq 2^2</math>. So we make it a <math>2^2</math> and solve <math>x-2^2=3</math>. We continue on using this same method where we move the perfect square up by <math>1</math> each time until we get the <math>x</math> to be a value greater than the RHS plus the perfect square number. When we repeat this until we have <math>8</math> rows, we get:
+We repeat the same procedure except with <math>x-1^2=1</math> for the next row and <math>x-1^2=2</math> for the row after that. However, at the fourth row, we see that solving <math>x-1^2=3</math> yields <math>x=4</math>, in which case it would be incorrect since <math>1^2=1</math> is not the greatest perfect square less than or equal to <math>x</math> . So we make it a <math>2^2</math> and solve <math>x-2^2=3</math>. We continue on using this same method where we increase the perfect square until <math>x</math> can be made bigger than it. When we repeat this until we have <math>8</math> rows, we get:
-<cmath> \begin{align*} 7223\\ 7223-84^2 &= 167\\ 167-12^2 &= 23\\ 23-4^2 &= 7\\ 7-2^2 &= 3\\ 3-1^2 &= 2\\ 2-1^2 &= 1\\ 1-1^2 &= 0\end{align*} </cmath>
+<cmath> \begin{array}{ccccc}{}&{}&{}&{}&7223\\ 7223&-&84^{2}&=&167\\ 167&-&12^{2}&=&23\\ 23&-&4^{2}&=&7\\ 7&-&2^{2}&=&3\\ 3&-&1^{2}&=&2\\ 2&-&1^{2}&=&1\\ 1&-&1^{2}&=&0\\ \end{array} </cmath>
-Hence the solution is the last digit of <math>7223</math>, which is <math>3</math>.
+Hence the solution is the last digit of <math>7223</math>, which is <math>\boxed{\textbf{(B)}\ 3}</math>.

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Difference between revisions of "2010 AMC 10A Problems/Problem 25"

Revision as of 18:22, 29 January 2011