Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
 
(One intermediate revision by the same user not shown)
Line 4: Line 4:
 
<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>.
+
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 the elements of one of the sets <math>\{1,1,1,1\},\{1,1,2\}</math>, <math>\{2,2\}</math>, <math>\{1,3\}</math>, or <math>\{4\}</math>.
 +
 
 +
In the first case, <math>N = 1111 = 101 \cdot 11</math> so this is a counter example.
 +
 
 +
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.  
 +
 
 +
In the third case, <math>N=22</math> is excluded for being even.
 +
 
 +
In the fourth case, both <math>N=13</math> and <math>N=31</math> are prime.
 +
 
 +
In the last case <math>N=4</math> is excluded for being even.
 +
 
 +
This gives two counterexamples and the answer is <math>\fbox{C}</math>
 +
 
 +
==See Also==
 +
 
 +
{{AHSME box|year=1994|num-b=20|num-a=22}}
 +
{{MAA Notice}}

Latest revision as of 13:11, 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 the elements of one 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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_21&oldid=154390"