Difference between revisions of "2003 AMC 12B Problems/Problem 8"
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− | + | {{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #8]] and [[2003 AMC 10B Problems|2003 AMC 10B #13]]}} | |
+ | ==Problem== | ||
− | < | + | Let <math>\clubsuit(x)</math> denote the sum of the digits of the positive integer <math>x</math>. For example, <math>\clubsuit(8)=8</math> and <math>\clubsuit(123)=1+2+3=6</math>. For how many two-digit values of <math>x</math> is <math>\clubsuit(\clubsuit(x))=3</math>? |
− | + | <math>\textbf{(A) } 3 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 9 \qquad\textbf{(E) } 10 </math> | |
− | + | ==Solution== | |
+ | Let <math>y=\clubsuit (x)</math>. Since <math>x \leq 99</math>, we have <math>y \leq 18</math>. Thus if <math>\clubsuit (y)=3</math>, then <math>y=3</math> or <math>y=12</math>. The 3 values of <math>x</math> for which <math>\clubsuit (x)=3</math> are 12, 21, and 30, and the 7 values of <math>x</math> for which <math>\clubsuit (x)=12</math> are 39, 48, 57, 66, 75, 84, and 93. There are <math>\boxed{E}</math> = <math>\boxed{10}</math> values in all. | ||
+ | |||
+ | ~Andrew_Lu/Mismatchedcubing | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC12 box|year=2003|ab=B|num-b=7|num-a=9}} | ||
+ | {{AMC10 box|year=2003|ab=B|num-b=12|num-a=14}} | ||
+ | {{MAA Notice}} |
Latest revision as of 11:40, 1 August 2024
- The following problem is from both the 2003 AMC 12B #8 and 2003 AMC 10B #13, so both problems redirect to this page.
Problem
Let denote the sum of the digits of the positive integer
. For example,
and
. For how many two-digit values of
is
?
Solution
Let . Since
, we have
. Thus if
, then
or
. The 3 values of
for which
are 12, 21, and 30, and the 7 values of
for which
are 39, 48, 57, 66, 75, 84, and 93. There are
=
values in all.
~Andrew_Lu/Mismatchedcubing
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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 | |
All AMC 12 Problems and Solutions |
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.