Difference between revisions of "1994 AHSME Problems/Problem 21"

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<math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4 </math>
 
<math> \textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4 </math>
 
==Solution==
 
==Solution==
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Since the sum of the digits of <math>N</math> is <math>4</math> and none of the digits are <math>0</math>, <math>N</math>'s digits must be the elements of the sets <math>\{1,1,1,1\},\{1,1,2\},</math> or <math>\{1,3\}</math>. In the first case, the only possible <math>N</math> is <math>1111</math>, and it can be checked that this is a counterexample because it is divisible by <math>11</math>. In the second case, <math>N</math> is either <math>211</math> or <math>121</math>. It can be checked that <math>211</math> is indeed prime, while <math>121</math> is divisible by <math>11</math>. Finally in the third case, both <math>13,31</math> are prime. So the final answer is <math>\boxed{\textbf{(C)} 2}</math>.

Revision as of 12:28, 15 February 2016

Problem

Find the number of counter examples to the statement: \[``\text{If  N is an odd positive integer the sum of whose digits is 4 and none of whose digits is 0, then N is prime}."\] $\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$

Solution

Since the sum of the digits of $N$ is $4$ and none of the digits are $0$, $N$'s digits must be the elements of the sets $\{1,1,1,1\},\{1,1,2\},$ or $\{1,3\}$. In the first case, the only possible $N$ is $1111$, and it can be checked that this is a counterexample because it is divisible by $11$. In the second case, $N$ is either $211$ or $121$. It can be checked that $211$ is indeed prime, while $121$ is divisible by $11$. Finally in the third case, both $13,31$ are prime. So the final answer is $\boxed{\textbf{(C)} 2}$.