2007 AMC 12A Problems/Problem 22

Revision as of 23:42, 30 August 2020 by Alexlikemath (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
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 badly

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, so this solution would only take approximately 3-5 minutes, helpful in a contest.

Solution 5 (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.

Solution 6

Rearranging, we get 2007 - S(n) - S(S(n)) = n.

We can now try S(n) = 1-28, to find values of n. Then, you need to check whether S(n) = sum digits of n. If so, you found a solution!

The bash involves making a table with 3 columns: S(n), n, S(n). We write the numbers 1-28 in the S(n) column, we can find n using the above equation, and the third column we can just find the sum of the digits of n, and if the first column/3rd column match, we have a solution.

You will in the end find S(n) = 3, 18, 21, 24 yield solutions, which corrispond to n = 2001, 1980, 1983, 1977. There are 4 solutions, so the answer is $\boxed{\text{D}}$

-Alexlikemath

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS