Difference between revisions of "1996 AHSME Problems/Problem 12"
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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>. | 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>. | ||
+ | |||
+ | ==Solution 2 (rigorous but easy)== | ||
+ | |||
+ | \textdollar{}f(f(f(k))) = 27\textdollar{} | ||
+ | |||
+ | We will alternate k between even and odd. | ||
+ | |||
+ | If k is odd, then \textdollar{}f(k) in terms of k = (k+3)/2 | ||
+ | |||
+ | Next, we have \textdollar{}f((k+3)/2) and \textdollar{}(k+3)/2 is even so \textdollar{}f((k+3)/2)= (k+3)/4 \textdollar{} | ||
+ | |||
+ | We are given that \textdollar{}f((k+3)/4)= 27\textdollar{} | ||
+ | |||
+ | \textdollar{}(k+3)/4= 27\textdollar{} so \textdollar{}k= 105\textdollar{}. The sum of it's digits is 1+0+5= \textdollar{}6\textdollar{} | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=11|num-a=13}} | {{AHSME box|year=1996|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:12, 30 June 2024
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
.
Solution 2 (rigorous but easy)
\textdollar{}f(f(f(k))) = 27\textdollar{}
We will alternate k between even and odd.
If k is odd, then \textdollar{}f(k) in terms of k = (k+3)/2
Next, we have \textdollar{}f((k+3)/2) and \textdollar{}(k+3)/2 is even so \textdollar{}f((k+3)/2)= (k+3)/4 \textdollar{}
We are given that \textdollar{}f((k+3)/4)= 27\textdollar{}
\textdollar{}(k+3)/4= 27\textdollar{} so \textdollar{}k= 105\textdollar{}. The sum of it's digits is 1+0+5= \textdollar{}6\textdollar{}
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 |
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