Difference between revisions of "2007 AMC 12A Problems/Problem 22"
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You will in the end find S(n) = 3, 18, 21, 24 yield solutions, which corrispond to n = 2001, 1980, 1983, 1977. There are 4 solutions, so the answer is <math>\boxed{\text{D}}</math> | You will in the end find S(n) = 3, 18, 21, 24 yield solutions, which corrispond to n = 2001, 1980, 1983, 1977. There are 4 solutions, so the answer is <math>\boxed{\text{D}}</math> | ||
− | + | -Alexlikemath | |
== See also == | == See also == |
Revision as of 00:42, 31 August 2020
- The following problem is from both the 2007 AMC 12A #22 and 2007 AMC 10A #25, so both problems redirect to this page.
Contents
Problem
For each positive integer , let
denote the sum of the digits of
For how many values of
is
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.
Solution 5 (Rigorous)
Let the number of digits of be
. If
,
will already be greater than
. Notice that
is always at most
. Then if
,
will be at most
,
will be at most
, and
will be even smaller than
. Clearly we cannot reach a sum of
, unless
(i.e.
has
digits).
Then, let be a four digit number in the form
. Then
.
is the sum of the digits of
. We can represent
as the sum of the tens digit and the ones digit of
. The tens digit in the form of a decimal is
.
To remove the decimal portion, we can simply take the floor of the expression,
.
Now that we have expressed the tens digit, we can express the ones digit as times the above expression, or
.
Adding the two expressions yields the value of
.
Combining this expression to the ones for and
yields
.
Setting this equal to and rearranging a bit yields
.
(The reason for this slightly weird arrangement will soon become evident)
Now we examine the possible values of . If
,
is already too large.
must also be greater than
, or
would be a
-digit number. Therefore,
. Now we examine by case.
If , then
and
must both be
(otherwise
would already be greater than
). Substituting these values into the equation yields
.
Sure enough, .
Now we move onto the case where . Then our initial equation simplifies to
Since and
can each be at most
, we substitute that value to find the lower bound of
. Doing so yields
.
The floor expression is at least , so the right-hand side is at least
. Solving for
, we see that
. Again, we substitute for
and the equation becomes
.
Just like we did for , we can find the lower bound of
by assuming
and solving:
The right hand side is for
and
for
. Solving for c yields
. Looking back at the previous equation, the floor expression is
for
and
for
. Thus, the right-hand side is
for
and
for
. We can solve these two scenarios as systems of equations/inequalities:
and
Solving yields three pairs
;
; and
. Checking the numbers
,
, and
; we find that all three work. Therefore there are a total of
possibilities for
.
Note: Although this solution takes a while to read (as well as to write) the actual time it takes to think through the process above is very short in comparison to the solution length.
Solution 6
Rearranging, we get 2007 - S(n) - S(S(n)) = n.
We can now try S(n) = 1-28, to find values of n. Then, you need to check whether S(n) = sum digits of n. If so, you found a solution!
The bash involves making a table with 3 columns: S(n), n, S(n). We write the numbers 1-28 in the S(n) column, we can find n using the above equation, and the third column we can just find the sum of the digits of n, and if the first column/3rd column match, we have a solution.
You will in the end find S(n) = 3, 18, 21, 24 yield solutions, which corrispond to n = 2001, 1980, 1983, 1977. There are 4 solutions, so the answer is
-Alexlikemath
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.