Difference between revisions of "2007 AMC 12A Problems/Problem 22"

Revision as of 14:30, 6 November 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 $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $n + S(n) + S(S(n)) = 2007?$

$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$

Solution

Solution 1

For the sake of notation let $T(n) = n + S(n) + S(S(n))$ . Obviously $n<2007$ . Then the maximum value of $S(n) + S(S(n))$ is when $n = 1999$ , and the sum becomes $28 + 10 = 38$ . So the minimum bound is $1969$ . We do casework upon the tens digit:

Case 1: $196u \Longrightarrow u = 9$ . Easy to directly disprove.

Case 2: $197u$ . $S(n) = 1 + 9 + 7 + u = 17 + u$ , and $S(S(n)) = 8+u$ if $u \le 2$ and $S(S(n)) = 2 + (u-3) = u-1$ otherwise.

Subcase a: $T(n) = 1970 + u + 17 + u + 8 + u = 1995 + 3u = 2007 \Longrightarrow u = 4$ . This exceeds our bounds, so no solution here.

Subcase b: $T(n) = 1970 + u + 17 + u + u - 1 = 1986 + 3u = 2007 \Longrightarrow u = 7$ . First solution.

Case 3: $198u$ . $S(n) = 18 + u$ , and $S(S(n)) = 9 + u$ if $u \le 1$ and $2 + (u-2) = u$ otherwise.

Subcase a: $T(n) = 1980 + u + 18 + u + 9 + u = 2007 + 3u = 2007 \Longrightarrow u = 0$ . Second solution.

Subcase b: $T(n) = 1980 + u + 18 + u + u = 1998 + 3u = 2007 \Longrightarrow u = 3$ . Third solution.

Case 4: $199u$ . But $S(n) > 19$ , and $n + S(n)$ clearly sum to $> 2007$ .

Case 5: $200u$ . So $S(n) = 2 + u$ and $S(S(n)) = 2 + u$ (recall that $n < 2007$ ), and $2000 + u + 2 + u + 2 + u = 2004 + 3u = 2007 \Longrightarrow u = 1$ . Fourth solution.

In total we have $4 \mathrm{(D)}$ solutions, which are $1977, 1980, 1983,$ and $2001$ .

Solution 2

Clearly, $n > 1950$ . We can break this into three cases:

Case 1: $n \geq 2000$

Inspection gives $n = 2001$ .

Case 2: $n < 2000$ , $n = 19xy$ (not to be confused with $19*x*y$ ), $x + y < 10$

If you set up an equation, it reduces to

$4x + y = 32$

which has as its only solution satisfying the constraints $x = 8$ , $y = 0$ .

Case 3: $n < 2000$ , $n = 19xy$ , $x + y \geq 10$

This reduces to

$4x + y = 35$ . The only two solutions satisfying the constraints for this equation are $x = 7$ , $y = 7$ and $x = 8$ , $y = 3$ .

The solutions are thus $1977, 1980, 1983, 2001$ and the answer is $\mathrm{(D)}\ 4$ .

Solution 3

As in Solution 1, we note that $S(n)\leq 28$ and $S(S(n))\leq 10$ .
Obviously, $n\equiv S(n)\equiv S(S(n)) \pmod 9$ .
As $2007\equiv 0 \pmod 9$ , this means that $n\bmod 9 \in\{0,3,6\}$ , or equivalently that $n\equiv S(n)\equiv S(S(n))\equiv 0 \pmod 3$ .

Thus $S(S(n))\in\{3,6,9\}$ . For each possible $S(S(n))$ we get three possible $S(n)$ .
(E. g., if $S(S(n))=6$ , then $S(n)=x$ is a number such that $x\leq 28$ and $S(x)=6$ , therefore $S(n)\in\{6,15,24\}$ .)

For each of these nine possibilities we compute $n_{?}$ as $2007-S(n)-S(S(n))$ and check whether $S(n_{?})=S(n)$ .
We'll find out that out of the 9 cases, in 4 the value $n_{?}$ has the correct sum of digits.
This happens for $n_{?}\in \{ 1977, 1980, 1983, 2001 \}$ .

Solution 4

This solution is not a good solution, but is viable for in contest situations

Clearly $n\equiv S(n) \pmod 9$ . Thus, $n+S(n)+S(S(n))\equiv 0 \pmod 9 \implies n\equiv 0\pmod 3.$ Now we need a bound for $n$ . It is clear that the maximum for $S(n)=36$ (from $n=9999$ ) which means the maximum for $S(n)+S(S(n))$ is $45$ . This means that $n\geq 1962$ .

Warning: This is where you will cringe

Now check all multiples of $3$ from $1962$ to $2007$ and we find that only $n=1977, 1980, 1983, 2001$ work, so our answer is $\mathrm{(D)}\ 4$ .

Remark: this may seem time consuming, but in reality, calculating $n+S(n)+S(S(n))$ for $16$ values is actually very quick, and this solution would take up around 3-5 minutes in contest, which isn't bad for a problem 22/25.

@@ Line 64: / Line 64: @@
 ===Solution 4===
-*This solution is not a good solution, but is viable for in contest situations*
+*This solution is not a good solution, but is viable for in contest situations
 Clearly <math>n\equiv S(n) \pmod 9</math>. Thus, <cmath>n+S(n)+S(S(n))\equiv 0 \pmod 9 \implies n\equiv 0\pmod 3.</cmath>
 Now we need a bound for <math>n</math>. It is clear that the maximum for <math>S(n)=36</math> (from <math>n=9999</math>) which means the maximum for <math>S(n)+S(S(n))</math> is <math>45</math>. This means that <math>n\geq 1962</math>.
-*Warning: This is where you will cringe*
+*Warning: This is where you will cringe
 Now check all multiples of <math>3</math> from <math>1962</math> to <math>2007</math> and we find that only <math>n=1977, 1980, 1983, 2001</math> work, so our answer is <math>\mathrm{(D)}\  4</math>.
-Remark: this may seem time consuming, but in reality, calculating <math>n+S(n)+S(S(n))</math> for <math>16</math> values is actually very quick, and this solution would take up around 3-5 minutes in contest, which isn't bad for a problem 22.
+Remark: this may seem time consuming, but in reality, calculating <math>n+S(n)+S(S(n))</math> for <math>16</math> values is actually very quick, and this solution would take up around 3-5 minutes in contest, which isn't bad for a problem 22/25.
 == 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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