Difference between revisions of "2007 AMC 12A Problems/Problem 22"
Mathwiz0803 (talk | contribs) (→Solution 4 (This is not allowed)) |
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<code> | <code> | ||
def calculateSumOfDigits(number): | def calculateSumOfDigits(number): | ||
+ | |||
number = str(number) | number = str(number) | ||
+ | |||
digits = [] | digits = [] | ||
+ | |||
for i in range(len(number)): | for i in range(len(number)): | ||
+ | |||
digits.append(int(number[i])) | digits.append(int(number[i])) | ||
+ | |||
return sum(digits) | return sum(digits) | ||
+ | |||
for i in range(2007): | for i in range(2007): | ||
+ | |||
if(i+calculateSumOfDigits(i)+calculateSumOfDigits(calculateSumOfDigits(i)) == 2007): | if(i+calculateSumOfDigits(i)+calculateSumOfDigits(calculateSumOfDigits(i)) == 2007): | ||
+ | |||
print(i) | print(i) | ||
+ | |||
</code> | </code> | ||
Revision as of 14:03, 26 November 2016
- 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
[hide]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 is not allowed)
This is mainly for fun. We can create a python program to do this for us:
def calculateSumOfDigits(number):
number = str(number)
digits = []
for i in range(len(number)):
digits.append(int(number[i]))
return sum(digits)
for i in range(2007):
if(i+calculateSumOfDigits(i)+calculateSumOfDigits(calculateSumOfDigits(i)) == 2007):
print(i)
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.