Difference between revisions of "1972 IMO Problems/Problem 3"
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Math-shinai (talk | contribs) (→Solution 2) |
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<math>\lfloor{2a}\rfloor+\lfloor{2b}\rfloor=2(a_1+b_1)+\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor</math> | <math>\lfloor{2a}\rfloor+\lfloor{2b}\rfloor=2(a_1+b_1)+\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor</math> | ||
− | On the other hand, <math>\lfloor{a}\rfloor+\lfloor{b}\rfloor+\lfloor{a+b}\rfloor=a_1+b_1+(a_1+b_1)+\lfloor\{a+b\}\rfloor</math> | + | On the other hand, <math>\lfloor{a}\rfloor+\lfloor{b}\rfloor+\lfloor{a+b}\rfloor=a_1+b_1+(a_1+b_1)+\lfloor\{2(a+b)\}\rfloor</math> |
− | It is trivial that <math>\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor\ge\lfloor\{a+b\}\rfloor</math> | + | It is trivial that <math>\lfloor2\{a\}\rfloor+\lfloor2\{b\}\rfloor\ge\lfloor\{(a+b)\}\rfloor</math> (Traingle Inequality) |
Apply Lemma 2.1 to the problem: and we are pretty much done. | Apply Lemma 2.1 to the problem: and we are pretty much done. | ||
Note: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did. | Note: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did. |
Revision as of 11:43, 14 November 2019
Let and
be arbitrary non-negative integers. Prove that
is an integer. (
.)
Solution 1
Let . We intend to show that
is integral for all
. To start, we would like to find a recurrence relation for
.
First, let's look at :
Second, let's look at :
Combining,
.
Therefore, we have found the recurrence relation .
We can see that is integral because the RHS is just
, which we know to be integral for all
.
So, must be integral, and then
must be integral, etc.
By induction, is integral for all
.
Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln723.html
Solution 2
Let p be a prime, and n be an integer. Let be the largest positive integer
such that
WTS: For all primes ,
We know
Lemma 2.1: Let be real numbers. Then
Proof of Lemma 2.1: Let and
On the other hand,
It is trivial that (Traingle Inequality)
Apply Lemma 2.1 to the problem: and we are pretty much done.
Note: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did.