Difference between revisions of "2012 AMC 10B Problems/Problem 20"

Revision as of 21:19, 28 February 2012

Problem

Bernardo and Silvia play the following game. An integer between $0$ and $999$ , inclusive, is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last person who produces a number less than $1000$ . Let $N$ be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of $N$ ?

$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Revision as of 21:19, 28 February 2012 (view source) Klu2014 (talk \| contribs) ← Older edit		Revision as of 21:19, 28 February 2012 (view source) Klu2014 (talk \| contribs) Newer edit →
Line 6:		Line 6:

	==Solution==		==Solution==
−
−	Let's test each number starting from <math>1</math>, for our first case <math>1</math>, we start out with <math>1</math> and the number is then given to Bernardo. He will double the given number so in this case <math>1\times 2=2</math>. So now he gives this resulting number to Silvia and she adds <math>50</math> to the number. So we have <math>2+50=52</math>. Now this is then passed to Bernardo so he doubles this <math>52</math> and we have <math>52\times 2=104</math>. And now this is then given to Silvia so <math>104+50=154</math>. So she passes it to Bernardo and we have <math>154\times 2=308</math>. Silvia gets it and so we have <math>308+50=358</math>. Continuing this we have <math>358\times 2=716</math>. There is no need to continue with this because when Silvia adds <math>50</math>

Page

Toolbox

Search

Difference between revisions of "2012 AMC 10B Problems/Problem 20"

Revision as of 21:19, 28 February 2012

Problem

Solution