Difference between revisions of "1975 USAMO Problems/Problem 1"
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<cmath>[3x + y] + [3y + x] \le [[\frac{3a+b+4}{5}]] + [[\frac{3b+a+4}{5}]] = S.</cmath> | <cmath>[3x + y] + [3y + x] \le [[\frac{3a+b+4}{5}]] + [[\frac{3b+a+4}{5}]] = S.</cmath> | ||
− | We shall prove that <math>S \le a + b = T</math>; to do so, we list cases. Without loss of generality, let <math>a \le b</math>. Then, we find, for all 15 cases: | + | We shall prove that <math>S \le a + b = T</math>; to do so, we list cases. Without loss of generality, let <math>a \le b</math>. Because <math>a</math> and <math>b</math> are less than one, we have <math>a \le b \le 4.</math> Then, we find, for all 15 cases: |
<math>a = 0, b = 0 \rightarrow S = 0, T = 0.</math> | <math>a = 0, b = 0 \rightarrow S = 0, T = 0.</math> |
Revision as of 18:23, 16 August 2014
Contents
Problem
(a) Prove that
where and denotes the greatest integer (e.g., ).
(b) Using (a) or otherwise, prove that
is integral for all positive integral and .
Background Knowledge
Note: A complete proof for this problem may require these results, and preferably also their proofs.
If , then . This is the alternate definition of .
If , then . This is easily proved by contradiction or consideration of the contrapositive.
If is an integer, then . This is proved from considering that .
This is a known fact: the exponent of a prime in the prime factorization of is .
Key Lemma
Lemma: For any pair of non-negative real numbers and , the following holds:
Proof of Key Lemma
We shall first prove the lemma statement for . Then , and so we have to prove that
Let , for integers and . Then , and so and .
Define a new function, the ceiling function of x, to be the least integer greater than or equal to x. Also, define the trun-ceil function, to be the value of the ceiling function minus one. Thus, if is an integer, and otherwise. It is not difficult to verify that if and are real numbers with , then . (The only new thing we have to consider here is the case where is integral, which is trivial.)
Therefore,
We shall prove that ; to do so, we list cases. Without loss of generality, let . Because and are less than one, we have Then, we find, for all 15 cases:
Thus, we have proved for all x and y in the range ,
Now, we prove the lemma statement without restrictions on x and y. Let , and , where , the fractional part of x, is defined to be . Note that as a result. Substituting gives the equivalent inequality
But, because for any integer , this is obtained from simplifications following the adding of to both sides of
which we have already proved (as ). Thus, the lemma is proved.
How the Key Lemma Solves the Problem
Part (a) is a direct corollary of the lemma.
For part (b), consider an arbitrary prime . We have to prove the exponent of in
is non-negative, or equivalently that
But, the right-hand side minus the left-hand side of this inequality is which is the sum of non-negative terms by the Lemma. Thus, the inequality is proved, and so, by considering all primes , we deduce that the exponents of all primes in are non-negative. This proves the integrality of (i.e. is an integer).
See Also
1975 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.