Difference between revisions of "2004 AIME II Problems/Problem 14"

(Solution 3)
(Solution 3)
 
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Case 2 : <math>a = 8</math>
 
Case 2 : <math>a = 8</math>
  
We get <math>11b + c = 112 \implies b = 0,1,2,\cdots ,10</math> and <math>c = 112 , 101 , 90,\cdots ,2</math> From here we get different <math>n</math>'s as <math>24 + 112 , 24 + 101 , 24 + 94, \cdots ,24 + 22</math>  (remember <math>n = 3a + 2b + c</math> and if you have difficulty in understanding how we got <math>n = 24 + (\cdots)</math> then just put the values of <math>a,b,c</math> i am sure you will get it :) )
+
We get <math>11b + c = 112 \implies b = 0,1,2,\cdots ,10</math> and <math>c = 112 , 101 , 90,\cdots ,2</math> From here we get different <math>n</math>'s as <math>24 + 112 , 24 + 103 , 24 + 94, \cdots ,24 + 22</math>  (remember <math>n = 3a + 2b + c</math> and if you have difficulty in understanding how we got <math>n = 24 + (\cdots)</math> then just put the values of <math>a,b,c</math> i am sure you will get it :) )
  
 
Let's write the sequences of <math>n</math>'s in a compact form , <math>T_p = 24 + 22 + 9(p-1)</math> (This will be helpful later on)
 
Let's write the sequences of <math>n</math>'s in a compact form , <math>T_p = 24 + 22 + 9(p-1)</math> (This will be helpful later on)

Latest revision as of 13:36, 22 February 2021

Problem

Consider a string of $n$ $7$'s, $7777\cdots77,$ into which $+$ signs are inserted to produce an arithmetic expression. For example, $7+77+777+7+7=875$ could be obtained from eight $7$'s in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000$?

Solution 1

Suppose we require $a$ $7$s, $b$ $77$s, and $c$ $777$s to sum up to $7000$ ($a,b,c \ge 0$). Then $7a + 77b + 777c = 7000$, or dividing by $7$, $a + 11b + 111c = 1000$. Then the question is asking for the number of values of $n = a + 2b + 3c$.

Manipulating our equation, we have $a + 2b + 3c = n = 1000 - 9(b + 12c) \Longrightarrow 0 \le 9(b+12c) < 1000$. Thus the number of potential values of $n$ is the number of multiples of $9$ from $0$ to $1000$, or $112$.


However, we forgot to consider the condition that $a \ge 0$. For a solution set $(b,c): n=1000-9(b+12c)$, it is possible that $a = n-2b-3c < 0$ (for example, suppose we counted the solution set $(b,c) = (1,9) \Longrightarrow n = 19$, but substituting into our original equation we find that $a = -10$, so it is invalid). In particular, this invalidates the values of $n$ for which their only expressions in terms of $(b,c)$ fall into the inequality $9b + 108c < 1000 < 11b + 111c$.

For $1000 - n = 9k \le 9(7 \cdot 12 + 11) = 855$, we can express $k$ in terms of $(b,c): n \equiv b \pmod{12}, 0 \le b \le 11$ and $c = \frac{n-b}{12} \le 7$ (in other words, we take the greatest possible value of $c$, and then "fill in" the remainder by incrementing $b$). Then $11b + 111c \le 855 + 2b + 3c \le 855 + 2(11) + 3(7) = 898 < 1000$, so these values work.

Similarily, for $855 \le 9k \le 9(8 \cdot 12 + 10) = 954$, we can let $(b,c) = (k-8 \cdot 12,8)$, and the inequality $11b + 111c \le 954 + 2b + 3c \le 954 + 2(10) + 3(8) = 998 < 1000$. However, for $9k \ge 963 \Longrightarrow n \le 37$, we can no longer apply this approach.

So we now have to examine the numbers on an individual basis. For $9k = 972$, $(b,c) = (0,9)$ works. For $9k = 963, 981, 990, 999 \Longrightarrow n = 37, 19, 10, 1$, we find (using that respectively, $b = 11,9,10,11 + 12p$ for integers $p$) that their is no way to satisfy the inequality $11b + 111c < 1000$.

Thus, the answer is $112 - 4 = \boxed{108}$.



A note: Above, we formulated the solution in a forward manner (the last four paragraphs are devoted to showing that all the solutions we found worked except for the four cases pointed out; in a contest setting, we wouldn't need to be nearly as rigorous). A more natural manner of attacking the problem is to think of the process in reverse, namely seeing that $n \equiv 1 \pmod{9}$, and noting that small values of $n$ would not work.

Looking at the number $7000$, we obviously see the maximum number of $7's$: a string of $1000 \ 7's$. Then, we see that the minimum is $28 \ 7's: \ 777*9 + 7 = 7000$. The next step is to see by what interval the value of $n$ increases. Since $777$ is $3 \ 7's, \ 77*10 + 7$ is $21 \ 7's$, we can convert a $777$ into $77's$ and $7's$ and add $18$ to the value of $n$. Since we have $9 \ 777's$ to work with, this gives us $28,46,64,82,100,118,136,154,172,190 ( = 28 + 18n | 1\leq n\leq 9)$ as values for $n$. Since $77$ can be converted into $7*11$, we can add $9$ to $n$ by converting $77$ into $7's$. Our $n = 190$, which has $0 \ 777's \ 90 \ 77's \ 10 7's$. We therefore can add $9$ to $n \ 90$ times by doing this. All values of $n$ not covered by this can be dealt with with the $n = 46 \ (8 \ 777's \ 10 \ 77's \ 2 \ 7's)$ up to $190$.

Solution 2

To simplify, replace all the $7$’s with $1$’s. Because the sum is congruent to $n \pmod 9$ and \[1000 \equiv 1 \pmod 9 \implies n \equiv 1 \pmod 9\] Also, $n \leq 1000$. There are \[\left \lfloor \frac{1000}{9} \right \rfloor + 1 = 112 \textrm{ positive integers that satisfy both conditions i.e. } \{1, 10, 19, 28, 37, 46, . . . , 1000\}.\]

For $n = 1, 10, 19$, the greatest sum that is less than or equal to $1000$ is $6 \cdot 111 + 1 = 677 \implies 112-3 = 109$.


Thus $n \geq 28$ and let $S = \{28, 37, 46, . . . , 1000\}$.

Note that $n=28$ is possible because $9 \cdot 111+1 \cdot 1 = 1000$.

When $n = 37$, the greatest sum that is at most $1000$ is $8 \cdot 111+6\cdot 11+1 \cdot 1 = 955$.


All other elements of $S$ are possible because if any element $n$ of $S$ between $46$ and $991$ is possible, then $(n+ 9)$ must be too.


$\textrm{Case } 1:\text{ Sum has no } 11$’s

It must have at least one $1$. If it has exactly one $1$, there must be nine $111$’s and $n = 28$. Thus, for $n \geq 46$, the sum has more than one $1$, so it must have at least $1000 - 8 \cdot 111 = 112$ number of $1$’s. For $n \leq 1000$, at least one $111$. To show that if $n$ is possible, then $(n + 9)$ is possible, replace a $111$ with $1 + 1 + 1$, replace eleven $(1 + 1)$’s with eleven $11$’s, and include nine new $1$’s as $+1$’s. The sum remains $1000$.


$\textrm{Case } 2: \textrm{ Sum has at least one } 11$.

Replace an $11$ with $1 + 1$, and include nine new $1$’s as $+1$’s. Now note that $46$ is possible because $8 \cdot 111 + 10 \cdot 11 + 2 \cdot 1 = 1000$. Thus all elements of $S$ except $37$ are possible.


Thus there are $\boxed{108}$ possible values for $n$.

~phoenixfire

Solution 3

It's obvious that we cannot have any number $\ge 7777$ because $7777 > 7000$ so the max number that an occur is $777$

Let's say we have $a$ 777's , $b$ 77's and $c$ 7's

From here we get our required equation as $777a + 77b + 7c = 7000$

Now comes the main problem , one might think that if we find number of $(a,b,c)$ then we're done , but in reality we are over-counting our number of $n$'s. This is because $n$ is the total number of 7's and from our equation we'll get $n$ as $3a + 2b + c$ (because there are three 7's , two 7's and one 7)

The reason why we're over-counting is because , say $a_1 , b_1 , c_1$ be a solution of our original equation and $a_2 , b_2 , c_2$ be another solution of our original equation , then there can be a possibility that $3a_1 + 2b_1 + c_1 = 3a_2 + 2b_2 + c_2$ where $a_1 \neq a_2 , b_1 \neq b_2 , c_1 \neq c_2$ (example : $2 + 3 + 4 = 1 + 5 + 3$ but $2 \neq 1 , 3 \neq 5 , 4 \neq 3$

We know that $0 \le a \le 9$ , $0 \le b \le 90$ , $0 \le c \le 1000$ The bound on $a$ is easier to handle with , so lets start putting values on $a$ and calculate $b , c , n$ by making cases

Reduced equation : $111a + 11b + c = 1000$

Case 1 : $a = 9$

We get $11b + c = 1 \implies b = 0 , c = 1$ is our only solution thus only $\boxed{1}$ value of $n$

Case 2 : $a = 8$

We get $11b + c = 112 \implies b = 0,1,2,\cdots ,10$ and $c = 112 , 101 , 90,\cdots ,2$ From here we get different $n$'s as $24 + 112 , 24 + 103 , 24 + 94, \cdots ,24 + 22$ (remember $n = 3a + 2b + c$ and if you have difficulty in understanding how we got $n = 24 + (\cdots)$ then just put the values of $a,b,c$ i am sure you will get it :) )

Let's write the sequences of $n$'s in a compact form , $T_p = 24 + 22 + 9(p-1)$ (This will be helpful later on) There are $\boxed{11}$ values of $n$

Case 3 : $a = 7$

We get $11b + c = 223 \implies b=0,1,2, \cdots ,20$ and $c = 223,212,201, \cdots ,3$ From here we get different $n$'s as $21 + 223, 21 + 214 , 21 + 205, \cdots ,21 + 43$

Let's also write the sequences of $n$'s in a compact form , $T_q = 21 + 43 + 9(q-1)$

Now comes the major part , since we need to find distinct $n$'s so let's subtract the cases where we find common values , from the total number of values.

To do this we need to make $T_p = T_q \implies p - q = 2$ (after some calculations you'll get $p - q = 2$) . Now we know that $0 \le p \le 10$ and $0 \le q \le 20$ so we get $p$ as $10,9,\cdots,2$ and $q$ as $8,7,\cdots,0$ so there are 9 common solutions out of 21(diff values of $q$) total , so there are $\boxed{21 - 9}$ values of $n$

Case 4 : $a = 6$

We get $11b + c = 334 \implies b = 0,1,2, \cdots ,30$ and $c = 334,323,312, \cdots ,4$ From here we get the different $n$'s as $18 + 334,18 + 325,18 + 316, \cdots ,18 + 64$

Compact form of terms is $T_r = 18 + 64 + 9(r-1)$ Now let's repeat the process of eliminating common solutions (one thing to notice is that we've removed common solutions of case 2 from case 3 so if we check case 4 with case 3 we'll remove common solutions from all previous cases and hence we do not have to handle common solutions of case 4 with case 2,1 and in general case X with case X-2,X-3,...2,1 , {basically means checking case X with case X-1 is enough})

$T_q = T_r \implies 21 + 43 + 9(q-1) = 18 + 64 + 9(r-1) \implies q - r = 2$ $\implies q = 20,19,18, \cdots ,2$ and $r = 18,17,16, \cdots ,0$ so there are 19 common solutions out of 31 (diff values of r) so there are $\boxed{31 - 19}$ values of $n$

Now hopefully you've seen a pattern , if not try another case 5 with $a = 5$ you'll get $\boxed{41 - 29}$ values of $n$

Like this we get the value of "distinct" $n$ by summing all the sub-values from the cases that we've solved.

$n = 1 + 11 + (21 - 9) + (31 - 19) + (41 - 29) + \cdots + (91 - 79)$

$\implies n = (1 + 11 + 21 + 31 + 41 + \cdots + 91) - (0 + 0 + 9 + 19 + 29 + \cdots + 79)$

$\implies \boxed{n = 108}$

~MrOreoJuice

See also

2004 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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