Difference between revisions of "2016 AMC 12A Problems/Problem 25"

Latest revision as of 17:28, 16 November 2022

Problem

Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k = 1$ , then the numbers that Bernardo writes are $16, 25, 36, 49, 64$ , and the numbers showing on the board after Silvia erases are $1, 2, 3, 4,$ and $6$ , and thus $f(1) = 5$ . What is the sum of the digits of $f(2) + f(4)+ f(6) + \dots + f(2016)$ ?

$\textbf{(A)}\ 7986\qquad\textbf{(B)}\ 8002\qquad\textbf{(C)}\ 8030\qquad\textbf{(D)}\ 8048\qquad\textbf{(E)}\ 8064$

Solution

Consider $f(2)$ . The numbers left on the blackboard will show the hundreds place at the end. In order for the hundreds place to differ by 2, the difference between two perfect squares needs to be at least $100$ . Since $100\le (x+1)^2-x^2=2x+1$ , this first happens at $x\ge \lfloor 99/2\rfloor = 50$ . The perfect squares from here go: $2500, 2601, 2704, 2809\dots$ . Note that the ones and tens also make the perfect squares, $1^2,2^2,3^2\dots$ . After the ones and tens make $100$ , the hundreds place will go up by $2$ , thus reaching our goal. Since $10^2=100$ , the last perfect square to be written will be $\left(50+10\right)^2=60^2=3600$ . The missing number is one less than the number of hundreds $(k=2)$ of $3600$ , or $35$ .

Now consider $f(4)$ . Instead of the difference between two squares needing to be $100$ , the difference must now be $10000$ . This first happens at $x\ge 5000$ . After this point, similarly, $\sqrt{10000}=100$ more numbers are needed to make the $10^4$ th's place go up by $2$ . This will take place at $\left(5000+100\right)^2=5100^2= 26010000$ . Removing the last four digits (the zeros) and subtracting one yields $2600$ for the skipped value.

In general, each new value of $f(k+2)$ will add two digits to the " $5$ " and one digit to the " $1$ ". This means that the last number Bernardo writes for $k=6$ is $\left(500000+1000\right)^2$ , the last for $k = 8$ will be $\left(50000000+10000\right)^2$ , and so on until $k=2016$ . Removing the last $k$ digits as Silvia does will be the same as removing $k/2$ trailing zeroes on the number to be squared. This means that the last number on the board for $k=6$ is $5001^2$ , $k=8$ is $50001^2$ , and so on. So the first missing number is $5001^2-1,50001^2-1\text{ etc.}$ The squaring will make a " $25$ " with two more digits than the last number, a " $10$ " with one more digit, and a " $1$ ". The missing number is one less than that, so the "1" will be subtracted from $f(k)$ . In other words, $f(k) = 25\cdot 10^{k-2}+1\cdot 10^{k/2}$ .

Therefore:

$f(2) =35 =25 +10$ $f(4) =2600 =2500 +100$ $f(6) =251000 =250000 +1000$ $f(8) = 25010000 = 25000000 + 10000$

And so on. The sum $f(2) + f(4) + f(6) +\dots + f(2016)$ is:

$2.52525252525\dots 2525\cdot 10^{2015}$ + $1.11111\dots 110\cdot 10^{1008}$ , with $2016$ repetitions each of " $25$ " and " $1$ ". There is no carrying in this addition. Therefore each $f(k)$ adds $2 + 5 + 1 = 8$ to the sum of the digits. Since $2n = 2016$ , $n = 1008$ , and $8n = 8064$ , or $\boxed{\textbf{(E)}\text{ 8064}}$ .

Solution 2 (Rigorous)

We assume $n \geq 1$ for all claims.

We will let $g_k(x) = \left \lfloor \frac{x^2}{10^k}\right \rfloor$ . This is the result when the last k digits are truncated off $x^2$ .

Let $x_n$ = the smallest a, such that $g_{2n}(a) - g_{2n}(a-1) \geq 2$

We then have $\left \lfloor \frac{a^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(a-1)^2}{10^{2n}} \right \rfloor \geq 2$

Claim 1: $x_n > 5\cdot10^{2n-1}$

Proof of Claim 1:

Assume for the sake of contradiction, we have $x_n \leq 5 \cdot 10^{2n-1}$

Let $c = \frac{2x_n - 1}{10^{2n}}$ , $d = \frac{(x_n-1)^2}{10^{2n}}$

Note that since $x_n \leq 5 \cdot 10^{2n-1}$ , we have $c < 1$

It is well known that $\lfloor i+j \rfloor \leq \lfloor i \rfloor + \lfloor j \rfloor + 1$ for any $i$ and $j$

Since $x_n$ satisfies the condition, we have:

$1 = \lfloor c \rfloor + 1 \geq \lfloor{c+d} \rfloor - \lfloor d \rfloor \geq 2$

This is a contradiction. $\blacksquare$

Claim 2: $x_n = 5 \cdot 10^{2n-1} + 10^n$

Proof of Claim 2:

We will show our choice of $x_n = 5 \cdot 10^{2n-1} + 10^n$ satisfies the criteria above.

$\left \lfloor \frac{{x_n}^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(x_n-1)^2}{10^{2n}} \right \rfloor = \left \lfloor \frac{25 \cdot 10^{4n - 2} + 10^{3n} + 10^{2n}}{10^{2n}} \right \rfloor - \left \lfloor \frac{25 \cdot 10^{4n - 2} + 10^{3n} - 2 \cdot 10^n + 1}{10^{2n}} \right \rfloor$

$= (25 \cdot 10^{2n-2} + 10^{n} + 1) - (25 \cdot 10^{2n-2} + 10^{n} - 1) = 2$

Now, we will show all values smaller than $5 \cdot 10^{2n-1} + 10^n$ don’t satisfy the criteria. We will assume for the sake of contradiction, there exists an $x'_n$ which satisfies the criteria.

We will write $x'_n$ as $5 \cdot 10^{2n - 1} + k$ . By our hypothesis, we have $k < 10^n$ . We also have $k > 0$ by claim 1.

We have $\left \lfloor \frac{{x'}_n^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{(x'_n-1)^2}{10^{2n}} \right \rfloor = \left \lfloor \frac{25 \cdot 10^{4n - 2} + k \cdot 10^{2n} + k^2}{10^{2n}} \right \rfloor - \left \lfloor \frac{25 \cdot 10^{4n - 2} + (k-1) \cdot 10^{2n} + (k-1)^2 }{10^{2n}} \right \rfloor$

$= (25 \cdot 10^{2n-2} + k) - (25 \cdot 10^{2n-2} + k - 1) = 1 \geq 2.$ This gets a contradiction. $\blacksquare$

Claim 3: $f(2n) = 25 \cdot10^{2n - 2} + 10^{n}$ .

Proof of Claim 3: Because of claim 2, $x_n$ is $5 \cdot 10^{2n-1} + 10^n$ . Note that $g_n(x_n)$ and $g_n(x_n - 1)$ are the first numbers written that will differ by at least 2. Thus, $f(2n) = g_n(x_n - 1) + 1 = 25 \cdot 10^{2n-2} + 10^n$ $\blacksquare$

Thus, $f(2) + f(4) \dots f(2016) = \underbrace{(2525 \dots 25)}_{1008 \ 25s} + \underbrace{(111111 \dots 1111)}_{1008 \ 1s}$ . This addition has no regroups/carry overs, so we can just take the sums of the digits of each of the addends, and sum them together to get 8064. $\boxed{(E)}$

-Alexlikemath

Video Solution by The Art of Problem Solving

https://www.youtube.com/watch?v=xVKNTQ9bKcg&list=PLyhPcpM8aMvI7N78mYZyatqveRU30iNcf&index=5

- AMBRIGGS

@@ Line 82: / Line 82: @@
 -Alexlikemath
-==Solution 3 (Cheap Realization)==
-If you are one of those people who are willing to take educated guesses, then just realize that <math>\textbf{(E)}\text{ 8064}</math> is the only answer choice that is a multiple of <math>2016</math>.
 ==Video Solution by The Art of Problem Solving==
@@ Line 93: / Line 89: @@
 - AMBRIGGS
-==See Also==
+==See Also=={{AMC12 box|year=2016|ab=A|num-b=24|after=Last Problem}}
-Related Question: [[2013 AIME II Problems/Problem 6]]
-{{AMC12 box|year=2016|ab=A|num-b=24|after=Last Problem}}
 {{MAA Notice}}

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
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

Page

Toolbox

Search