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

Revision as of 20:00, 14 September 2007

Problem

For each positive integer $n$ , let $S(n)$ denote the sum of the digits of $n.$ For how many values of $n$ is $\displaystyle n + S(n) + S(S(n)) = 2007?$

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

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 $\displaystyle S(n) + S(S(n))$ 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 $\displaystyle S(S(n)) = 2 + (u-3) = u-1$ 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: $\displaystyle T(n) = 1970 + u + 17 + u + u - 1 = 1986 + 3u = 2007 \Longrightarrow \displaystyle u = 7$ . First solution.

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

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

Case 5: $200u$ . So $S(n) = 2 + u$ and $S(S(n)) = 2 + u$ , and $\displaystyle 2000 + u + 2 + u + 2 + u = 2004 + 3u = 2007 \Longrightarrow u = 1$ . 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 $\displaystyle n = 2001$ .

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

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

This reduces to

$\displaystyle4x + y = 35$ . 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 $\mathrm{(D)}\ 4$ .

@@ Line 48: / Line 48: @@
 :<math>\displaystyle4x + y = 35</math>. The only two solutions satisfying the constraints for this equation are <math>x = 7</math>, <math>y = 7</math> and <math>x = 8</math>, <math>y = 3</math>.
-The solutions are thus <math>1977, 1980, 1983, 2001</math> and the answer is <math>D) 4</math>.
+The solutions are thus <math>1977, 1980, 1983, 2001</math> and the answer is <math>\mathrm{(D)}\  4</math>.
 == 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

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