Difference between revisions of "1996 AHSME Problems/Problem 12"
Talkinaway (talk | contribs) |
(→Solution 2 (rigorous but easy)) |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 32: | Line 32: | ||
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)== | ||
+ | |||
+ | <cmath> f(f(f(k))) = 27 </cmath> | ||
+ | |||
+ | We will work from the inside to the outside and alternate \( k \) between even and odd. | ||
+ | |||
+ | If \( k \) is odd, then \( f(k) \) in terms of \( k \) is \(\frac{k+3}{2}\). | ||
+ | |||
+ | Next, we have \( f\left(\frac{k+3}{2}\right) \), and \(\frac{k+3}{2}\) is even, so \( f\left(\frac{k+3}{2}\right) = \frac{k+3}{4} \). | ||
+ | |||
+ | We are given that \( f\left(\frac{k+3}{4}\right) = 27 \). | ||
+ | |||
+ | <cmath>\frac{k+3}{4} = 27 \implies k = 105.</cmath> | ||
+ | |||
+ | The sum of its digits is \( 1 + 0 + 5 = 6 \). | ||
+ | |||
+ | |||
+ | |||
+ | {gnv12} | ||
==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}} |
Revision as of 20:16, 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)
We will work from the inside to the outside and alternate \( k \) between even and odd.
If \( k \) is odd, then \( f(k) \) in terms of \( k \) is \(\frac{k+3}{2}\).
Next, we have \( f\left(\frac{k+3}{2}\right) \), and \(\frac{k+3}{2}\) is even, so \( f\left(\frac{k+3}{2}\right) = \frac{k+3}{4} \).
We are given that \( f\left(\frac{k+3}{4}\right) = 27 \).
The sum of its digits is \( 1 + 0 + 5 = 6 \).
{gnv12}
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.