1991 AIME Problems/Problem 6
Contents
Problem
Suppose is a real number for which
Find . (For real , is the greatest integer less than or equal to .)
Solution (Hopefully Intuitive)
Note that the value of up to the closest multiple of doesn't matter, so assume is an integer. By Hermite's Identity, this equation is equivalent to
We can guess that is about 600-something. Assuming that , (so all the floors after 92 are 7 but all the floors before 18 is 6) this equation becomes Solving, we get that which is not in the range that we guessed. Oops. This means that we need for to actually go above since this equation assumes that the sum of the floors is smaller than they actually are once we surpass .
We now guess that . Similar to before, we solve to get which is in the range we guessed! So, the answer is .
~~solasky
Solution
There are numbers in the sequence. Since the terms of the sequence can be at most apart, all of the numbers in the sequence can take one of two possible values. Since , the values of each of the terms of the sequence must be either or . As the remainder is , must take on of the values, with being the value of the remaining numbers. The 39th number is , which is also the first term of this sequence with a value of , so . Solving shows that , so , and .
Solution 2 (Faster)
Recall by Hermite's Identity that for positive integers , and real . Similar to above, we quickly observe that the last 35 take the value of 8, and remaining first ones take a value of 7. So, and . We can see that . Because is at most 7, and is at least 8, we can clearly see their values are and respectively. So, , and . Since there are 19 terms in the former equation and 8 terms in the latter, our answer is
Note
In the contest, you would just observe this mentally, and then calculate , hence the speed at which one can carry out this solution.
See also
1991 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 7 | |
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