2007 AMC 12A Problems/Problem 22
- 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 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
Claim. The only positive integers that satisfy the condition are perfect multiples of
.
Proof of claim:
We examine the positive integers mod . Here are the cases.
Case 1. . Now, we examine
modulo
.
Case 1.1. The tens digit of
is different from the tens digit of the largest multiple of
under
. (In other words, this means we will carry when adding from the perfect multiple of
under
.)
Observe that when we carry, i.e. Add
onto
to obtain
, the units digit decreases by
while the tens digit increases by
. This means that the sum of the digits decreases by
in total, and we have
, so the "mod 9" of the sum increases by
. This means that, regardless of whether the sum carries or not, the modulo 9 of the sum of the digits always increases by
.
Case 1.2. The tens digits are the same, which is trivial since the units digit just increases by which means that the sum is also equivalent to
.
This means that and similarly letting the next
,
. Summing these, we have
. Clearly, no integers of this form will satisfy the condition because
is a perfect multiple of
.
Case 2. .
In this case, we apply exactly the same argument. There is at most one carry, which means that the sum of the digits will always be congruent to mod
. Then we can apply similar arguments to get
and
, so adding gives
.
It is trivial to see that for , for
, we must have
. Only when
is
a multiple of
, which means that
must be a multiple of
.
Now, we find the integers. Again, consider two cases: Integers that are direct multiples of and integers that are multiples of
but not
.
Case 1. is a multiple of
. An integer of the form
will not work since the least such integer is
which already exceeds our bounds. Thus, we need only consider the integers of the form
. The valid sums of the digits of
are
and
in this case.
Case 1.1. The sum of the digits is . This means that
, so
. Clearly this number satisfies our constraints.
Case 1.2. The sum of the digits is . This means that
, ,so
. Since the sum of the digits of
is not
, this does not work.
This means that there is integer in this case.
Case 2. is a multiple of
, not
.
.
Case 2.1. Integers of the form
. Then
or
; it is trivial to see that
exceeds our bounds, so
and
.
Case 2.2. Integers of the form . Then
and we consider each case separately.
Case 2.2.1. Integers with . That means
which clearly does not work.
Case 2.2.2. Integers with . That means
which also does not work
Case 2.2.3. Integers with . That means
which is valid.
Case 2.2.4. Integers with . That means
which is also valid.
We have considered every case, so there are integers that satisfy the given condition.
~Refined by HamstPan38825
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
.
We can now try the integers from to
inclusive for
to find values of
. Then, you need to check whether
equals the sum of the digits of
. If so, you have found a solution!
The bash involves making a table with columns:
. We write the numbers
in the
column, we can find
using the above equation, and the third column we can just find the sum of the digits of
, and if the first column and 3rd column match, we have a solution.
You will in the end find yield solutions, which correspond to
. There are
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 |
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