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

Revision as of 20:32, 3 July 2013

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 the these 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, $n > 1900$ . We can break this up into three cases:

Case 1: $n \geq 2000$

Inspection gives $n = 2001$ .

Case 2: $n < 2000$ , $n = 19xy$ , $x + y < 10$

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 $n_{?}\in \{ 1977, 1980, 1983, 2001 \}$ .

Revision as of 21:57, 3 July 2009 (view source) Mathemagician1729 (talk \| contribs) m (→‎Solution 1) ← Older edit		Revision as of 20:32, 3 July 2013 (view source) Nathan wailes (talk \| contribs) (→‎See also) Newer edit →
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	[[Category:Introductory Number Theory Problems]]		[[Category:Introductory Number Theory Problems]]
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2007 AMC 12A (Problems • Answer Key • Resources)
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