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==
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>.
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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>, <math>{2,2}</math>, <math>\{1,3\}</math>, or <math>\{4\}</math>.
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In the first case, <math>N = 1111 = 101 \cdot 11</math> so this is a counter example.
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In the second case, <math>N=112</math> is excluded for being even.  With <math>N=121=11^2</math> we have a counterexample.  We can check <math>N=211</math> by trial division, and verify it is indeed prime.  
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In the third case, <math>N=22</math> is excluded for being even.
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In the fourth case, both <math>N=13</math> and <math>N=31</math> are prime.
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In the last case <math>N=4</math> is excluded for being even.
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This gives two counterexamples and the answer is <math>\fbox{C}</math>
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==See Also==
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{{AHSME box|year=1994|num-b=20|num-a=22}}
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{{MAA Notice}}

Revision as of 03:23, 28 May 2021

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 $\{1,1,1,1\},\{1,1,2\}$, ${2,2}$, $\{1,3\}$, or $\{4\}$.

In the first case, $N = 1111 = 101 \cdot 11$ so this is a counter example.

In the second case, $N=112$ is excluded for being even. With $N=121=11^2$ we have a counterexample. We can check $N=211$ by trial division, and verify it is indeed prime.

In the third case, $N=22$ is excluded for being even.

In the fourth case, both $N=13$ and $N=31$ are prime.

In the last case $N=4$ is excluded for being even.

This gives two counterexamples and the answer is $\fbox{C}$

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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 26 27 28 29 30
All AHSME Problems and Solutions

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