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

Latest revision as of 22:08, 28 July 2024

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 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 $2007-38=1969$ . 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: $n < 2000$ , $n = \overline{19xy}$

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: $n < 2000$ , $n = \overline{19xy}$ , $x + y \geq 10$

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 (Fastest casework)

It is well-known that $n \equiv S(n)\equiv S(S(n)) \pmod{9}.$ Substituting, we have that $n+n+n \equiv 2007 \pmod{9} \implies n \equiv 0 \pmod{3}.$ Since $n \leq 2007,$ we must have that $\max S(n)=1+9+9+9=28.$ Now, we list out the possible values for $S(n)$ in a table, noting that it is a multiple of $3$ because $n$ is a multiple of $3.$

$\begin{tabular}{c|c c c c c c c c c c} S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\ \end{tabular}$

Then, we compute the corresponding values of $S(S(n)).$

$\begin{tabular}{c|c c c c c c c c c c} S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\ \hline S(S(n)) & 0 & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\ \end{tabular}$

Finally, we may compute the corresponding values of $n$ using the fact that $n=2007-S(n)-S(S(n)).$

$\begin{tabular}{c|c c c c c c c c c c} n & 2007 & 2001 & 1995 & 1989 & 1992 & 1986 & 1980 & 1983 & 1977 & 1971 \\ \hline S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\ \hline S(S(n)) & 0 & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\ \end{tabular}$

Notice how all conditions are designed to be satisfied except whether $S(n)$ is accurate with respect to $n.$ So, the only thing that remains is to check this. We may eliminate, for example, when $n=2007$ we have $S(n)=9$ while the table states that it is $0.$ Proceeding similarly, we obtain the following table.

$\begin{tabular}{c|c c c c c c c c c c} n & \cancel{2007} & \boxed{2001} & \cancel{1995} & \cancel{1989} & \cancel{1992} & \cancel{1986} & \boxed{1980} & \boxed{1983} & \boxed{1977} & \cancel{1971} \\ \hline S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\ \hline S(S(n)) & 0 & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\ \end{tabular}$

It follows that there are $\boxed{4}\implies \textbf{(D)}$ possible values for $n.$ ~samrocksnature

Solution 4

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 5

Claim. The only positive integers $n$ that satisfy the condition are perfect multiples of $3$ .

Proof of claim: We examine the positive integers mod $9$ . Here are the cases.

Case 1. $n \equiv 1 \pmod 9$ . Now, we examine $S(n)$ modulo $9$ . Case 1.1. The tens digit of $n$ is different from the tens digit of the largest multiple of $9$ under $n$ . (In other words, this means we will carry when adding from the perfect multiple of $9$ under $n$ .) Observe that when we carry, i.e. Add $1$ onto $1989$ to obtain $1990$ , the units digit decreases by $9$ while the tens digit increases by $1$ . This means that the sum of the digits decreases by $8$ in total, and we have $-8 \equiv 1 \pmod 9$ , so the "mod 9" of the sum increases by $1$ . This means that, regardless of whether the sum carries or not, the modulo 9 of the sum of the digits always increases by $1$ .

Case 1.2. The tens digits are the same, which is trivial since the units digit just increases by $1$ which means that the sum is also equivalent to $1 \pmod 9$ .

This means that $S(n) \equiv 1 \pmod 9$ and similarly letting the next $n=S(n)$ , $S(S(n)) \equiv 1 \pmod 9$ . Summing these, we have $n+S(n)+S(S(n)) \equiv 3 \pmod 9$ . Clearly, no integers of this form will satisfy the condition because $2007$ is a perfect multiple of $9$ .

Case 2. $n \equiv 2 \pmod 9$ .

In this case, we apply exactly the same argument. There is at most one carry, which means that the sum of the digits will always be congruent to $2$ mod $9$ . Then we can apply similar arguments to get $S(n) \equiv 2 \pmod 9$ and $S(S(n)) \equiv 2 \pmod 9$ , so adding gives $n+S(n)+S(S(n)) \equiv 6 \pmod 9$ .

It is trivial to see that for $n \equiv k \pmod 9$ , for $0 \leq k \leq 8$ , we must have $n+S(n)+S(S(n)) \equiv 3k \pmod 9$ . Only when $k=0, 3, 6$ is $3k$ a multiple of $9$ , which means that $n$ must be a multiple of $3$ .

Now, we find the integers. Again, consider two cases: Integers that are direct multiples of $9$ and integers that are multiples of $3$ but not $9$ .

Case 1. $n$ is a multiple of $9$ . An integer of the form $\overline{20ab}$ will not work since the least such integer is $2007$ which already exceeds our bounds. Thus, we need only consider the integers of the form $\overline{19ab}$ . The valid sums of the digits of $n$ are $18$ and $27$ in this case.

Case 1.1. The sum of the digits is $18$ . This means that $S(n)=18, S(S(n))=9$ , so $n=2007-18-9=1980$ . Clearly this number satisfies our constraints.

Case 1.2. The sum of the digits is $27$ . This means that $S(n)=27, S(S(n))=9$ , ,so $n=2007-27-9=1971$ . Since the sum of the digits of $1971$ is not $27$ , this does not work.

This means that there is $1$ integer in this case.

Case 2. $n$ is a multiple of $3$ , not $9$ . . Case 2.1. Integers of the form $\overline{20ab}$ . Then $S(n)=3$ or $S(n)=6$ ; it is trivial to see that $S(n)=6$ exceeds our bounds, so $S(n)=3$ and $n=2007-6=2001$ .

Case 2.2. Integers of the form $\overline{19ab}$ . Then $S(n)=12, 15, 21, 24$ and we consider each case separately.

Case 2.2.1. Integers with $S(n)=12$ . That means $n=2007-12-3=1992$ which clearly does not work.

Case 2.2.2. Integers with $S(n)=15$ . That means $n=2007-15-6=1986$ which also does not work

Case 2.2.3. Integers with $S(n)=21$ . That means $n=2007-21-3=1983$ which is valid.

Case 2.2.4. Integers with $S(n)=24$ . That means $n=2007-24-6=1977$ which is also valid.

We have considered every case, so there are $\boxed{4}$ integers that satisfy the given condition.

~Refined by HamstPan38825

Solution 6 (Rigorous)

Let the number of digits of $n$ be $m$ . If $m = 5$ , $n$ will already be greater than $2007$ . Notice that $S(n)$ is always at most $9m$ . Then if $m = 3$ , $n$ will be at most $999$ , $S(n)$ will be at most $27$ , and $S(S(n))$ will be even smaller than $27$ . Clearly we cannot reach a sum of $2007$ , unless $m = 4$ (i.e. $n$ has $4$ digits).

Then, let $n$ be a four digit number in the form $1000a + 100b + 10c + d$ . Then $S(n) = a + b + c + d$ .

$S(S(n))$ is the sum of the digits of $a + b + c + d$ . We can represent $S(S(n))$ as the sum of the tens digit and the ones digit of $S(n)$ . The tens digit in the form of a decimal is

$\frac{a + b + c + d}{10}$ .

To remove the decimal portion, we can simply take the floor of the expression,

$\lfloor\frac{a + b + c + d}{10}\rfloor$ .

Now that we have expressed the tens digit, we can express the ones digit as $S(n) -10$ times the above expression, or

$a + b + c + d - 10\lfloor\frac{a + b + c + d}{10}\rfloor$ .

Adding the two expressions yields the value of $S(S(n))$

$= a + b + c + d - 9\lfloor\frac{a + b + c + d}{10}\rfloor$ .

Combining this expression to the ones for $n$ and $S(n)$ yields

$1002a + 102b + 12c + 3d - 9\lfloor\frac{a + b + c + d}{10}\rfloor$ .

Setting this equal to $2007$ and rearranging a bit yields

$12c + 3d = 2007 - 1002a - 102b + 9\lfloor\frac{a + b + c + d}{10}\rfloor$

$\Rightarrow$ $4c + d = 669 - 334a - 34b + 3\lfloor\frac{a + b + c + d}{10}\rfloor$ .

(The reason for this slightly weird arrangement will soon become evident)

Now we examine the possible values of $a$ . If $a \ge 3$ , $n$ is already too large. $a$ must also be greater than $0$ , or $n$ would be a $3$ -digit number. Therefore, $a = 1 \, \text{or} \, 2$ . Now we examine by case.

If $a = 2$ , then $b$ and $c$ must both be $0$ (otherwise $n$ would already be greater than $2007$ ). Substituting these values into the equation yields

$d = 1 + 3\lfloor\frac{2 + d}{10}\rfloor$

$\Rightarrow$ $d=1$ .

Sure enough, $2001 + (2+1) + 3=2007$ .

Now we move onto the case where $a = 1$ . Then our initial equation simplifies to

$4c + d = 335 - 34b + 3\lfloor\frac{1 + b + c + d}{10}\rfloor$

Since $c$ and $d$ can each be at most $9$ , we substitute that value to find the lower bound of $b$ . Doing so yields

$34b \ge 290 + 3\lfloor\frac{19 + b}{10}\rfloor$ .

The floor expression is at least $3\lfloor\frac{19}{10}\rfloor=3$ , so the right-hand side is at least $293$ . Solving for $b$ , we see that $b \ge 9$ $\Rightarrow$ $b=9$ . Again, we substitute for $b$ and the equation becomes

$4c + d = 29 + 3\lfloor\frac{10 + c + d}{10}\rfloor$

$\Rightarrow$ $4c + d = 32 + 3\lfloor\frac{c + d}{10}\rfloor$ .

Just like we did for $b$ , we can find the lower bound of $c$ by assuming $d = 9$ and solving:

$4c + 9 \ge 29 + 3\lfloor\frac{c + 9}{10}\rfloor$

$\Rightarrow$ $4c \ge 20 + 3\lfloor\frac{c + 9}{10}\rfloor$

The right hand side is $20$ for $c=0$ and $23$ for $c \ge 1$ . Solving for c yields $c \ge 6$ . Looking back at the previous equation, the floor expression is $0$ for $c+d \le 9$ and $3$ for $c+d \ge 10$ . Thus, the right-hand side is $32$ for $c+d \le 9$ and $35$ for $c+d \ge 10$ . We can solve these two scenarios as systems of equations/inequalities:

$4c+d = 32$

$c+d \le 9$

and

$4c+d=35$

$c+d \ge 10$

Solving yields three pairs $(c, d):$ $(8, 0)$ ; $(8, 3)$ ; and $(7, 7)$ . Checking the numbers $1980$ , $1983$ , and $1977$ ; we find that all three work. Therefore there are a total of $4$ possibilities for $n$ $\Rightarrow$ $\boxed{\text{D}}$ .

Note: Although this solution takes a while to read (as well as to write) the actual time it takes to think through the process above is very short in comparison to the solution length.

~edits by vadava_lx

@@ Line 5: / Line 5: @@
 <math>\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5</math>
-== Solution ==
+== Solution 1 ==
-=== Solution 1 ===
+For the sake of notation, let <math>T(n) = n + S(n) + S(S(n))</math>. Obviously <math>n<2007</math>. Then the maximum value of <math>S(n) + S(S(n))</math> is when <math>n = 1999</math>, and the sum becomes <math>28 + 10 = 38</math>. So the minimum bound is <math>2007-38=1969</math>. We do [[casework]] upon the tens digit:
-For the sake of notation let <math>T(n) = n + S(n) + S(S(n))</math>. Obviously <math>n<2007</math>. Then the maximum value of <math>S(n) + S(S(n))</math> is when <math>n = 1999</math>, and the sum becomes <math>28 + 10 = 38</math>. So the minimum bound is <math>1969</math>. We do [[casework]] upon the tens digit:
 Case 1: <math>196u \Longrightarrow u = 9</math>. Easy to directly disprove.
@@ Line 27: / Line 26: @@
 In total we have <math>4 \mathrm{(D)}</math> solutions, which are <math>1977, 1980, 1983, </math> and <math>2001</math>.
-=== Solution 2 ===
+== Solution 2 ==
 Clearly, <math>n > 1950</math>. We can break this into three cases:
@@ Line 34: / Line 33: @@
 :Inspection gives <math>n = 2001</math>.
-Case 2: <math>n < 2000</math>, <math>n = 19xy</math> (not to be confused with <math>19*x*y</math>), <math>x + y < 10 </math>
+Case 2: <math>n < 2000</math>, <math>n = \overline{19xy}</math>
 :If you set up an equation, it reduces to
@@ Line 42: / Line 41: @@
 :which has as its only solution satisfying the constraints <math>x = 8</math>, <math>y = 0</math>.
-Case 3: <math>n < 2000</math>, <math>n = 19xy</math>, <math>x + y \geq 10</math>
+Case 3: <math>n < 2000</math>, <math>n = \overline{19xy}</math>, <math>x + y \geq 10</math>
 :This reduces to
@@ Line 50: / Line 49: @@
 The solutions are thus <math>1977, 1980, 1983, 2001</math> and the answer is <math>\mathrm{(D)}\  4</math>.
-=== Solution 3 ===
+== Solution 3 (Fastest casework)==
+It is well-known that <math>n \equiv S(n)\equiv S(S(n)) \pmod{9}.</math> Substituting, we have that <cmath>n+n+n \equiv 2007 \pmod{9} \implies n \equiv 0 \pmod{3}.</cmath> Since <math>n \leq 2007,</math> we must have that <math>\max S(n)=1+9+9+9=28.</math> Now, we list out the possible values for <math>S(n)</math> in a table, noting that it is a multiple of <math>3</math> because <math>n</math> is a multiple of <math>3.</math>
+<center>
+<math>\begin{tabular}{c|c c c c c c c c c c}
+ S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\
+\end{tabular}</math>
+</center>
+Then, we compute the corresponding values of <math>S(S(n)).</math>
+<center>
+<math>\begin{tabular}{c|c c c c c c c c c c}
+ S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\
+\hline
+S(S(n)) & 0 & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\
+\end{tabular}</math>
+</center>
+Finally, we may compute the corresponding values of <math>n</math> using the fact that <math>n=2007-S(n)-S(S(n)).</math>
+<center>
+<math>\begin{tabular}{c|c c c c c c c c c c}
+n & 2007 & 2001 & 1995 & 1989 & 1992 & 1986 & 1980 & 1983 & 1977 & 1971 \\
+\hline
+S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\
+\hline
+S(S(n)) & 0 & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\
+\end{tabular}</math>
+</center>
+Notice how all conditions are designed to be satisfied except whether <math>S(n)</math> is accurate with respect to <math>n.</math> So, the only thing that remains is to check this. We may eliminate, for example, when <math>n=2007</math> we have <math>S(n)=9</math> while the table states that it is <math>0.</math> Proceeding similarly, we obtain the following table.
+<center>
+<math>\begin{tabular}{c|c c c c c c c c c c}
+n & \cancel{2007} & \boxed{2001} & \cancel{1995} & \cancel{1989} & \cancel{1992} & \cancel{1986} & \boxed{1980} & \boxed{1983} & \boxed{1977} & \cancel{1971} \\
+\hline
+ S(n) & 0 & 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24 & 27 \\
+\hline
+S(S(n)) & 0 & 3 & 6 & 9 & 3 & 6 & 9 & 3 & 6 & 9 \\
+\end{tabular}</math>
+</center>
+It follows that there are <math>\boxed{4}\implies \textbf{(D)}</math> possible values for <math>n.</math> ~samrocksnature
+== Solution 4 ==
 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/>
@@ Line 62: / Line 106: @@
 This happens for <math>n_{?}\in \{ 1977, 1980, 1983, 2001 \}</math>.
-===Solution 4===
+==Solution 5==
-*This solution is not a good solution, but is viable for in contest situations
+Claim. The only positive integers <math>n</math> that satisfy the condition are perfect multiples of <math>3</math>.
-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>.
+Proof of claim:
-*Warning: This is where you will cringe badly
+We examine the positive integers mod <math>9</math>. Here are the cases.
-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>.
+Case 1. <math>n \equiv 1 \pmod 9</math>. Now, we examine <math>S(n)</math> modulo <math>9</math>.
+Case 1.1. The tens digit of <math>n</math> is different from the tens digit of the largest multiple of <math>9</math> under <math>n</math>. (In other words, this means we will carry when adding from the perfect multiple of <math>9</math> under <math>n</math>.)
+Observe that when we carry, i.e. Add <math>1</math> onto <math>1989</math> to obtain <math>1990</math>, the units digit decreases by <math>9</math> while the tens digit increases by <math>1</math>. This means that the sum of the digits decreases by <math>8</math> in total, and we have <math>-8 \equiv 1 \pmod 9</math>, so the "mod 9" of the sum increases by <math>1</math>. This means that, regardless of whether the sum carries or not, the modulo 9 of the sum of the digits always increases by <math>1</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, so this solution would only take approximately 3-5 minutes, helpful in a contest.
+Case 1.2. The tens digits are the same, which is trivial since the units digit just increases by <math>1</math> which means that the sum is also equivalent to <math>1 \pmod 9</math>.
-==Solution 5 (Rigorous)==
+This means that <math>S(n) \equiv 1 \pmod 9</math> and similarly letting the next <math>n=S(n)</math>, <math>S(S(n)) \equiv 1 \pmod 9</math>. Summing these, we have <math>n+S(n)+S(S(n)) \equiv 3 \pmod 9</math>. Clearly, no integers of this form will satisfy the condition because <math>2007</math> is a perfect multiple of <math>9</math>.
+Case 2. <math>n \equiv 2 \pmod 9</math>.
+In this case, we apply exactly the same argument. There is at most one carry, which means that the sum of the digits will always be congruent to <math>2</math> mod <math>9</math>. Then we can apply similar arguments to get <math>S(n) \equiv 2 \pmod 9</math> and <math>S(S(n)) \equiv 2 \pmod 9</math>, so adding gives <math>n+S(n)+S(S(n)) \equiv 6 \pmod 9</math>.
+It is trivial to see that for <math>n \equiv k \pmod 9</math>, for <math>0 \leq k \leq 8</math>, we must have <math>n+S(n)+S(S(n)) \equiv 3k \pmod 9</math>. Only when <math>k=0, 3, 6</math> is <math>3k</math> a multiple of <math>9</math>, which means that <math>n</math> must be a multiple of <math>3</math>.
+Now, we find the integers. Again, consider two cases: Integers that are direct multiples of <math>9</math> and integers that are multiples of <math>3</math> but not <math>9</math>.
+Case 1. <math>n</math> is a multiple of <math>9</math>. An integer of the form <math>\overline{20ab}</math> will not work since the least such integer is <math>2007</math> which already exceeds our bounds. Thus, we need only consider the integers of the form <math>\overline{19ab}</math>. The valid sums of the digits of <math>n</math> are <math>18</math> and <math>27</math> in this case.
+Case 1.1. The sum of the digits is <math>18</math>. This means that <math>S(n)=18, S(S(n))=9</math>, so <math>n=2007-18-9=1980</math>. Clearly this number satisfies our constraints.
+Case 1.2. The sum of the digits is <math>27</math>. This means that <math>S(n)=27, S(S(n))=9</math>, ,so <math>n=2007-27-9=1971</math>. Since the sum of the digits of <math>1971</math> is not <math>27</math>, this does not work.
+This means that there is <math>1</math> integer in this case.
+Case 2. <math>n</math> is a multiple of <math>3</math>, not <math>9</math>.
+.
+Case 2.1. Integers of the form <math>\overline{20ab}</math>. Then <math>S(n)=3</math> or <math>S(n)=6</math>; it is trivial to see that <math>S(n)=6</math> exceeds our bounds, so <math>S(n)=3</math> and <math>n=2007-6=2001</math>.
+Case 2.2. Integers of the form <math>\overline{19ab}</math>. Then <math>S(n)=12, 15, 21, 24</math> and we consider each case separately.
+Case 2.2.1. Integers with <math>S(n)=12</math>. That means <math>n=2007-12-3=1992</math> which clearly does not work.
+Case 2.2.2. Integers with <math>S(n)=15</math>. That means <math>n=2007-15-6=1986</math> which also does not work
+Case 2.2.3. Integers with <math>S(n)=21</math>. That means <math>n=2007-21-3=1983</math> which is valid.
+Case 2.2.4. Integers with <math>S(n)=24</math>. That means <math>n=2007-24-6=1977</math> which is also valid.
+We have considered every case, so there are <math>\boxed{4}</math> integers that satisfy the given condition.
+~Refined by HamstPan38825
+==Solution 6 (Rigorous)==
 Let the number of digits of <math>n</math> be <math>m</math>. If <math>m = 5</math>, <math>n</math> will already be greater than <math>2007</math>. Notice that <math>S(n)</math> is always at most <math>9m</math>. Then if <math>m = 3</math>, <math>n</math> will be at most <math>999</math>, <math>S(n)</math> will be at most <math>27</math>, and <math>S(S(n))</math> will be even smaller than <math>27</math>. Clearly we cannot reach a sum of <math>2007</math>, unless <math>m = 4</math> (i.e. <math>n</math> has <math>4</math> digits).
@@ Line 88: / Line 171: @@
-Now that we have expressed the tens digit, we can express the ones digit as <math>S(S(n)) -10</math> times the above expression, or
+Now that we have expressed the tens digit, we can express the ones digit as <math>S(n) -10</math> times the above expression, or
@@ Line 172: / Line 255: @@
 Note: Although this solution takes a while to read (as well as to write) the actual time it takes to think through the process above is very short in comparison to the solution length.
+~edits by vadava_lx
 == See also ==

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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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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