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

Revision as of 14:52, 16 August 2014

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 will 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 $x, y < 1$ . 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 $5x < a + 1 and 5y < b + 1$ , 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, $[3x + y] + [3y + x] \le [[\frac{3a+b+4}{5}]] + [[\frac{3b+a+4}{5}]] = S.$

To prove that $S \le a + b = T$ , we shall list cases. Without loss of generality, let $a \le b$ . (There are only 15 cases, all simple computations: $T$ is obvious, and $S$ is easy. Compare this to an assembly line.)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 collorary 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

\[\sum_{k=1}^{\infty} \left( \left[ \frac{5m}{p^k} \right] + \left[ \frac{5n}{p^k} \right] - \left( \left[ \frac{m}{p^k} \right] + \left[ \frac{n}{p^k} \right] + \left[ \frac{3m+n}{p^k} \right] + \left[ \frac{3n+m}{p^k} \right] \right),\] (Error compiling LaTeX. Unknown error_msg)

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 12: / Line 12: @@
 ''Note: A complete proof for this problem will require these results, and preferably also their proofs.''
-If <math>[x] = a</math>, then <math>x < a + 1</math>. This is the definition of greatest integer less than itself.
+If <math>[x] = a</math>, then <math>a \le x < a + 1</math>. This is the alternate definition of <math>[x]</math>.
 If <math>a < b</math>, then <math>[a] \le [b]</math>. This is easily proved by contradiction or consideration of the contrapositive.
@@ Line 19: / Line 19: @@
 This is a known fact: the exponent of a prime <math>p</math> in the prime factorization of <math>n!</math> is <math>\sum_{k=1}^\infty \left[ \frac{n}{p^k} \right]</math>.
 ==Key Lemma==

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