Difference between revisions of "1991 AIME Problems/Problem 6"
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Find <math>\lfloor 100r \rfloor</math>. (For real <math>x^{}_{}</math>, <math>\lfloor x \rfloor</math> is the [[floor function|greatest integer]] less than or equal to <math>x^{}_{}</math>.) | Find <math>\lfloor 100r \rfloor</math>. (For real <math>x^{}_{}</math>, <math>\lfloor x \rfloor</math> is the [[floor function|greatest integer]] less than or equal to <math>x^{}_{}</math>.) | ||
+ | |||
+ | ==Solution (Hopefully Intuitive)== | ||
+ | Note that the value of <math>r</math> up to the closest multiple of <math>\frac{1}{100}</math> doesn't matter, so assume <math>100r</math> is an integer. By Hermite's Identity, this equation is equivalent to <cmath>\left\lfloor 100r \right\rfloor - \left(\left\lfloor r + \frac{1}{100}\right\rfloor + \cdots + \left\lfloor r + \frac{18}{100}\right\rfloor + \left\lfloor r + \frac{92}{100}\right\rfloor + \cdots + \left\lfloor r + \frac{99}{100}\right\rfloor \right)</cmath> <cmath>=546.</cmath> | ||
+ | We can guess that <math>100r</math> is about 600-something. Assuming that <math>608 \le 100r < 682</math>, (so all the floors after 92 are 7 but all the floors before 18 is 6) this equation becomes <cmath>100r - 6 \cdot 18 - 7 \cdot 8 = 546.</cmath> Solving, we get that <cmath>100r = 708,</cmath> which is not in the range that we guessed. Oops. This means that we need for <math>100r</math> to actually go above <math>708</math> since this equation assumes that the sum of the floors is smaller than they actually are once we surpass <math>682</math>. | ||
+ | |||
+ | We now guess that <math>708 \le 100r < 782</math>. Similar to before, we solve <cmath>100r - 7 \cdot 18 - 8 \cdot 8 = 546</cmath> to get <cmath>100r = 736,</cmath> which is in the range we guessed! So, the answer is <math>736</math>. | ||
+ | |||
+ | ~~solasky | ||
== Solution == | == Solution == |
Revision as of 14:49, 25 July 2023
Contents
[hide]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 thatis 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) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.