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

Revision as of 23:42, 30 August 2020

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

@@ Line 185: / Line 185: @@
 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 <math>\boxed{\text{D}}</math>
+-Alexlikemath
 == 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

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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