Difference between revisions of "2007 AMC 12A Problems/Problem 22"
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Clearly <math>n\equiv S(n) \pmod 9</math>. Thus, <cmath>n+S(n)+S(S(n))\equiv 0 \pmod 9 \implies n\equiv 0\pmod 3.</cmath> | Clearly <math>n\equiv S(n) \pmod 9</math>. Thus, <cmath>n+S(n)+S(S(n))\equiv 0 \pmod 9 \implies n\equiv 0\pmod 3.</cmath> | ||
Now we need a bound for <math>n</math>. It is clear that the maximum for <math>S(n)=36</math> (from <math>n=9999</math>) which means the maximum for <math>S(n)+S(S(n))</math> is <math>45</math>. This means that <math>n\geq 1962</math>. | Now we need a bound for <math>n</math>. It is clear that the maximum for <math>S(n)=36</math> (from <math>n=9999</math>) which means the maximum for <math>S(n)+S(S(n))</math> is <math>45</math>. This means that <math>n\geq 1962</math>. | ||
− | *Warning: This is where you will cringe | + | *Warning: This is where you will cringe badly |
Now check all multiples of <math>3</math> from <math>1962</math> to <math>2007</math> and we find that only <math>n=1977, 1980, 1983, 2001</math> work, so our answer is <math>\mathrm{(D)}\ 4</math>. | Now check all multiples of <math>3</math> from <math>1962</math> to <math>2007</math> and we find that only <math>n=1977, 1980, 1983, 2001</math> work, so our answer is <math>\mathrm{(D)}\ 4</math>. | ||
− | Remark: this may seem time consuming, but in reality, calculating <math>n+S(n)+S(S(n))</math> for <math>16</math> values is actually very quick, | + | Remark: this may seem time consuming, but in reality, calculating <math>n+S(n)+S(S(n))</math> for <math>16</math> values is actually very quick, so this solution would only take approximately 3-5 minutes, helpful in a contest. |
== See also == | == See also == |
Revision as of 15:55, 26 July 2019
- The following problem is from both the 2007 AMC 12A #22 and 2007 AMC 10A #25, so both problems redirect to this page.
Problem
For each positive integer , let
denote the sum of the digits of
For how many values of
is
Contents
Solution
Solution 1
For the sake of notation let . Obviously
. Then the maximum value of
is when
, and the sum becomes
. So the minimum bound is
. We do casework upon the tens digit:
Case 1: . Easy to directly disprove.
Case 2: .
, and
if
and
otherwise.
- Subcase a:
. This exceeds our bounds, so no solution here.
- Subcase b:
. First solution.
Case 3: .
, and
if
and
otherwise.
- Subcase a:
. Second solution.
- Subcase b:
. Third solution.
Case 4: . But
, and
clearly sum to
.
Case 5: . So
and
(recall that
), and
. Fourth solution.
In total we have solutions, which are
and
.
Solution 2
Clearly, . We can break this into three cases:
Case 1:
- Inspection gives
.
Case 2: ,
(not to be confused with
),
- If you set up an equation, it reduces to
- which has as its only solution satisfying the constraints
,
.
Case 3: ,
,
- This reduces to
. The only two solutions satisfying the constraints for this equation are
,
and
,
.
The solutions are thus and the answer is
.
Solution 3
As in Solution 1, we note that and
.
Obviously, .
As , this means that
, or equivalently that
.
Thus . For each possible
we get three possible
.
(E. g., if , then
is a number such that
and
, therefore
.)
For each of these nine possibilities we compute as
and check whether
.
We'll find out that out of the 9 cases, in 4 the value has the correct sum of digits.
This happens for .
Solution 4
- This solution is not a good solution, but is viable for in contest situations
Clearly . Thus,
Now we need a bound for
. It is clear that the maximum for
(from
) which means the maximum for
is
. This means that
.
- Warning: This is where you will cringe badly
Now check all multiples of from
to
and we find that only
work, so our answer is
.
Remark: this may seem time consuming, but in reality, calculating for
values is actually very quick, so this solution would only take approximately 3-5 minutes, helpful in a contest.
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 21 |
Followed by Problem 23 |
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 |
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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.