Difference between revisions of "2007 AMC 12A Problems/Problem 22"
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The solutions are thus <math>1977, 1980, 1983, 2001</math> and the answer is <math>\mathrm{(D)}\ 4</math>. | The solutions are thus <math>1977, 1980, 1983, 2001</math> and the answer is <math>\mathrm{(D)}\ 4</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | As in Solution 1, we note that <math>S(n)\leq 28</math> and <math>S(S(n))\leq 10</math>. <br/> | ||
+ | Obviously, <math>n\equiv S(n)\equiv S(S(n)) \pmod 9</math>. <br/> | ||
+ | As <math>2007\equiv 0 \pmod 9</math>, this means that <math>n\bmod 9 \in\{0,3,6\}</math>, or equivalently that <math>n\equiv S(n)\equiv S(S(n))\equiv 0 \pmod 3</math>. | ||
+ | |||
+ | Thus <math>S(S(n))\in\{3,6,9\}</math>. For each possible <math>S(S(n))</math> we get three possible <math>S(n)</math>. <br/> | ||
+ | (E. g., if <math>S(S(n))=6</math>, then <math>S(n)=x</math> is a number such that <math>x\leq 28</math> and <math>S(x)=6</math>, therefore <math>S(n)\in\{6,15,24\}</math>.) | ||
+ | |||
+ | For each of these nine possibilities we compute <math>n_{?}</math> as <math>2007-S(n)-S(S(n))</math> and check whether <math>S(n_{?})=S(n)</math>. <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>. | ||
== See also == | == See also == |
Revision as of 08:47, 5 January 2009
- 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.
Case 3: . , and if and otherwise.
- Subcase a: . Second solution.
- Subcase b: . Third solution.
Case 4: . But , and the these clearly sum to .
Case 5: . So and , and . Fourth solution.
In total we have solutions, which are and .
Solution 2
Clearly, . We can break this up into three cases:
Case 1:
- Inspection gives .
Case 2: , ,
- 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 .
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 |