Difference between revisions of "2007 AMC 12A Problems/Problem 22"
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We'll find out that out of the 9 cases, in 4 the value <math>n_{?}</math> has the correct sum of digits. <br/> | We'll find out that out of the 9 cases, in 4 the value <math>n_{?}</math> has the correct sum of digits. <br/> | ||
This happens for <math>n_{?}\in \{ 1977, 1980, 1983, 2001 \}</math>. | This happens for <math>n_{?}\in \{ 1977, 1980, 1983, 2001 \}</math>. | ||
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== See also == | == See also == | ||
Revision as of 17:53, 12 June 2018
- 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 
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.
. This exceeds our bounds, so no solution here.
. First solution.Case 3: 
. 
, and 
if 
and 
otherwise.
- Subcase a:
. Second solution. - Subcase b:
. Third solution.
. Second solution.
. 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 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 
.
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. 
