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.
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 (PYTHON!)
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.