Difference between revisions of "1996 AHSME Problems/Problem 12"
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+ | ==Problem 12== | ||
+ | |||
+ | A function <math>f</math> from the integers to the integers is defined as follows: | ||
+ | |||
+ | <cmath> f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases} </cmath> | ||
+ | |||
+ | Suppose <math>k</math> is odd and <math>f(f(f(k))) = 27</math>. What is the sum of the digits of <math>k</math>? | ||
+ | |||
+ | <math> \textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15 </math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | ===First iteration=== | ||
+ | To get <math>f(k) = 27</math>, you could either have <math>f(27 - 3)</math> and add <math>3</math>, or <math>f(27\cdot 2)</math> and divide by <math>2</math>. | ||
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+ | If you had the former, you would have <math>f(24)</math>, and the function's rule would have you divide. Thus, <math>k=54</math> is the only number for which <math>f(k) = 27</math>. | ||
+ | |||
+ | ===Second iteration=== | ||
+ | Going out one step, if you have <math>f(f(k)) = 27</math>, you would have to have <math>f(k) = 54</math>. For <math>f(k) = 54</math>, you would either have <math>f(54-3)</math> and add <math>3</math>, or <math>f(54\cdot 2)</math> and divide by <math>2</math>. | ||
+ | |||
+ | Both are possible: <math>f(51)</math> and <math>f(108)</math> return values of <math>54</math>. Thus, <math>f(f(51)) = f(54) = 27</math>, and <math>f(f(108)) = f(54) = 27</math>. | ||
+ | |||
+ | |||
+ | ===Third iteration=== | ||
+ | Going out the final step, if you have <math>f(f(f(k))) = 27</math>, you would have to have <math>f(f(k))) = 51</math> or <math>f(f(k)) = 108</math>. | ||
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+ | If you doubled either of these, <math>k</math> would not be odd. So you must subtract <math>3</math>. | ||
+ | |||
+ | If you subtract <math>3</math> from <math>51</math>, you would compute <math>f(48)</math>, which would halve it, and not add the <math>3</math> back. | ||
+ | |||
+ | If you subtract <math>3</math> from <math>108</math>, you would compute <math>f(105)</math>, which would add the <math>3</math> back. | ||
+ | |||
+ | Thus, <math>f(f(f(105))) = f(f(108)) = f(54) = 27</math>, and <math>105</math> is odd. The desired sum of the digits is <math>6</math>, and the answer is <math>\boxed{B}</math>. | ||
+ | |||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=11|num-a=13}} | {{AHSME box|year=1996|num-b=11|num-a=13}} |
Revision as of 21:19, 18 August 2011
Contents
Problem 12
A function from the integers to the integers is defined as follows:
Suppose is odd and . What is the sum of the digits of ?
Solution
First iteration
To get , you could either have and add , or and divide by .
If you had the former, you would have , and the function's rule would have you divide. Thus, is the only number for which .
Second iteration
Going out one step, if you have , you would have to have . For , you would either have and add , or and divide by .
Both are possible: and return values of . Thus, , and .
Third iteration
Going out the final step, if you have , you would have to have or .
If you doubled either of these, would not be odd. So you must subtract .
If you subtract from , you would compute , which would halve it, and not add the back.
If you subtract from , you would compute , which would add the back.
Thus, , and is odd. The desired sum of the digits is , and the answer is .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 |