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 | + | 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 first case, <math>N = 1111 = 101 \cdot 11</math> so this is a counter example. |
Latest revision as of 14:11, 28 May 2021
Problem
Find the number of counter examples to the statement:
Solution
Since the sum of the digits of is and none of the digits are , 's digits must the elements of one of the sets , , , or .
In the first case, so this is a counter example.
In the second case, is excluded for being even. With we have a counterexample. We can check by trial division, and verify it is indeed prime.
In the third case, is excluded for being even.
In the fourth case, both and are prime.
In the last case is excluded for being even.
This gives two counterexamples and the answer is
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 |
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