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

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 $1969$. 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 vales 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 $S(S(n)) -10$ 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.