Difference between revisions of "2004 AIME II Problems/Problem 14"
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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 + | + | 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 's, into which signs are inserted to produce an arithmetic expression. For example, could be obtained from eight 's in this way. For how many values of is it possible to insert signs so that the resulting expression has value ?
Solution 1
Suppose we require s, s, and s to sum up to (). Then , or dividing by , . Then the question is asking for the number of values of .
Manipulating our equation, we have . Thus the number of potential values of is the number of multiples of from to , or .
However, we forgot to consider the condition that . For a solution set , it is possible that (for example, suppose we counted the solution set , but substituting into our original equation we find that , so it is invalid). In particular, this invalidates the values of for which their only expressions in terms of fall into the inequality .
For , we can express in terms of and (in other words, we take the greatest possible value of , and then "fill in" the remainder by incrementing ). Then , so these values work.
Similarily, for , we can let , and the inequality . However, for , we can no longer apply this approach.
So we now have to examine the numbers on an individual basis. For , works. For , we find (using that respectively, for integers ) that their is no way to satisfy the inequality .
Thus, the answer is .
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 , and noting that small values of would not work.
Looking at the number , we obviously see the maximum number of : a string of . Then, we see that the minimum is . The next step is to see by what interval the value of increases. Since is is , we can convert a into and and add to the value of . Since we have to work with, this gives us as values for . Since can be converted into , we can add to by converting into . Our , which has . We therefore can add to times by doing this. All values of not covered by this can be dealt with with the up to .
Solution 2
To simplify, replace all the ’s with ’s. Because the sum is congruent to and Also, . There are
For , the greatest sum that is less than or equal to is .
Thus and let .
Note that is possible because .
When , the greatest sum that is at most is .
All other elements of are possible because if any element of between and is possible, then must be too.
’s
It must have at least one . If it has exactly one , there must be nine ’s and . Thus, for , the sum has more than one , so it must have at least number of ’s. For , at least one . To show that if is possible, then is possible, replace a with , replace eleven ’s with eleven ’s, and include nine new ’s as ’s. The sum remains .
.
Replace an with , and include nine new ’s as ’s. Now note that is possible because . Thus all elements of except are possible.
Thus there are possible values for .
~phoenixfire
Solution 3
It's obvious that we cannot have any number because so the max number that an occur is
Let's say we have 777's , 77's and 7's
From here we get our required equation as
Now comes the main problem , one might think that if we find number of then we're done , but in reality we are over-counting our number of 's. This is because is the total number of 7's and from our equation we'll get as (because there are three 7's , two 7's and one 7)
The reason why we're over-counting is because , say be a solution of our original equation and be another solution of our original equation , then there can be a possibility that where (example : but
We know that , , The bound on is easier to handle with , so lets start putting values on and calculate by making cases
Reduced equation :
Case 1 :
We get is our only solution thus only value of
Case 2 :
We get and From here we get different 's as (remember and if you have difficulty in understanding how we got then just put the values of i am sure you will get it :) )
Let's write the sequences of 's in a compact form , (This will be helpful later on) There are values of
Case 3 :
We get and From here we get different 's as
Let's also write the sequences of 's in a compact form ,
Now comes the major part , since we need to find distinct '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 (after some calculations you'll get ) . Now we know that and so we get as and as so there are 9 common solutions out of 21(diff values of ) total , so there are values of
Case 4 :
We get and From here we get the different 's as
Compact form of terms is 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})
and so there are 19 common solutions out of 31 (diff values of r) so there are values of
Now hopefully you've seen a pattern , if not try another case 5 with you'll get values of
Like this we get the value of "distinct" by summing all the sub-values from the cases that we've solved.
~MrOreoJuice
See also
2004 AIME II (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.