Difference between revisions of "1975 USAMO Problems/Problem 1"

Revision as of 22:35, 16 November 2017

Problem

(a) Prove that

$[5x]+[5y]\ge [3x+y]+[3y+x]$ ,

where $x,y\ge 0$ and $[u]$ denotes the greatest integer $\le u$ (e.g., $[\sqrt{2}]=1$ ).

(b) Using (a) or otherwise, prove that

$\frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}$

is integral for all positive integral $m$ and $n$ .

Background Knowledge

Note: A complete proof for this problem may require these results, and preferably also their proofs.

If $[x] = a$ , then $a \le x < a + 1$ . This is the alternate definition of $[x]$ .

If $a < b$ , then $[a] \le [b]$ . This is easily proved by contradiction or consideration of the contrapositive.

If $a$ is an integer, then $[x + a] = [x] + a$ . This is proved from considering that $[x] + a \le x + a < [x] + a + 1$ .

This is a known fact: the exponent of a prime $p$ in the prime factorization of $n!$ is $\sum_{k=1}^\infty \left[ \frac{n}{p^k} \right]$ .

Key Lemma

Lemma: For any pair of non-negative real numbers $x$ and $y$ , the following holds: $[5x] + [5y] \ge [x] + [y] + [3x + y] + [3y + x].$

Proof of Key Lemma

We shall first prove the lemma statement for $0 \le x, y < 1$ . Then $[x] = [y] = 0$ , and so we have to prove that $[5x] + [5y] \ge [3x + y] + [3y + x].$

Let $[5x] = a, [5y] = b$ , for integers $a$ and $b$ . Then $5x < a + 1 \text{ and } 5y < b + 1$ , and so $x < \frac{a+1}{5}$ and $y < \frac{b+1}{5}$ .

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, $[[x]],$ to be the value of the ceiling function minus one. Thus, $[[a]] = a - 1$ if $a$ is an integer, and $[[a]] = [a]$ otherwise. It is not difficult to verify that if $a$ and $b$ are real numbers with $a < b$ , then $[[a]] \le [a] \le [[b]]$ . (The only new thing we have to consider here is the case where $b$ is integral, which is trivial.)

Therefore, $[3x + y] + [3y + x] \le \left [ \left [\frac{3a+b+4}{5} \right ] \right] + \left[\left[\frac{3b+a+4}{5} \right]\right] = S.$

We shall prove that $S \le a + b = T$ ; to do so, we list cases. Without loss of generality, let $a \le b$ . Because $x$ and $y$ are less than one, we have $a \le b \le 4.$ Then, we find, for all 15 cases:

$a = 0, b = 0 \rightarrow S = 0, T = 0.$

$a = 0, b = 1 \rightarrow S = 1, T = 1.$

$a = 0, b = 2 \rightarrow S = 2, T = 2.$

$a = 0, b = 3 \rightarrow S = 3, T = 3.$

$a = 0, b = 4 \rightarrow S = 4, T = 4.$

$a = 1, b = 1 \rightarrow S = 2, T = 2.$

$a = 1, b = 2 \rightarrow S = 3, T = 3.$

$a = 1, b = 3 \rightarrow S = 3, T = 4.$

$a = 1, b = 4 \rightarrow S = 5, T = 5.$

$a = 2, b = 2 \rightarrow S = 4, T = 4.$

$a = 2, b = 3 \rightarrow S = 4, T = 5.$

$a = 2, b = 4 \rightarrow S = 5, T = 6.$

$a = 3, b = 3 \rightarrow S = 6, T = 6.$

$a = 3, b = 4 \rightarrow S = 6, T = 7.$

$a = 4, b = 4 \rightarrow S = 6, T = 8.$

Thus, we have proved for all x and y in the range $[0, 1)$ , $[5x] + [5y] = T \ge S \ge [3x + y] + [3y + x].$

Now, we prove the lemma statement without restrictions on x and y. Let $x = [x] + \{x\}$ , and $y = [y] + \{y\}$ , where $\{x\}$ , the fractional part of x, is defined to be $x - [x]$ . Note that $\{x\} < 1$ as a result. Substituting gives the equivalent inequality

$[5[x] + 5\{x\}] + [5[y] + 5\{y\}] \ge [x] + [y] + [3[x] + 3\{x\} + [y] + \{y\}] + [3[y] + 3\{y\} + [x] + \{x\}].$

But, because $[x] + a = [x + a]$ for any integer $a$ , this is obtained from simplifications following the adding of $5[x] + 5[y]$ to both sides of

$[5\{x\}] + [5\{y\}] \ge [3\{x\} + \{y\}] + [3\{y\} + \{x\}],$

which we have already proved (as $0 \le \{x\}, \{y\} < 1$ ). 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 $p$ . We have to prove the exponent of $p$ in

$I = \frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}$

is non-negative, or equivalently that $\sum_{k=1}^{\infty} \left( \left[ \frac{5m}{p^k} \right] + \left[ \frac{5n}{p^k} \right] \right) \ge \sum_{k=1}^{\infty} \left( \left[ \frac{m}{p^k} \right] + \left[ \frac{n}{p^k} \right] + \left[ \frac{3m+n}{p^k} \right] + \left[ \frac{3n+m}{p^k} \right] \right).$

But, the right-hand side minus the left-hand side of this inequality is $\sum_{k=1}^{\infty} \left( \left[ \frac{5m}{p^k} \right] + \left[ \frac{5n}{p^k} \right] - \left[ \frac{m}{p^k} \right] - \left[ \frac{n}{p^k} \right] - \left[ \frac{3m+n}{p^k} \right] - \left[ \frac{3n+m}{p^k} \right] \right),$ which is the sum of non-negative terms by the Lemma. Thus, the inequality is proved, and so, by considering all primes $p$ , we deduce that the exponents of all primes in $I$ are non-negative. This proves the integrality of $I$ (i.e. $I$ is an integer). $\blacksquare$

@@ Line 32: / Line 32: @@
 Therefore,
-<cmath>[3x + y] + [3y + x] \le [[\frac{3a+b+4}{5}]] + [[\frac{3b+a+4}{5}]] = S.</cmath>
+<cmath>[3x + y] + [3y + x] \le \left [ \left [\frac{3a+b+4}{5} \right ] \right] + \left[\left[\frac{3b+a+4}{5} \right]\right] = 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>. Because <math>x</math> and <math>y</math> are less than one, we have <math>a \le b \le 4.</math> Then, we find, for all 15 cases:

1975 USAMO (Problems • Resources)
Preceded by First Question	Followed by Problem 2
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

Page

Toolbox

Search