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

## 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 $x < a + 1$. This is the definition of greatest integer less than itself.

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$is integral, which is trivial.) Therefore, <cmath>[3x + y] + [3y + x] \le [[\frac{3a+b+4}{5}]] + [[\frac{3b+a+4}{5}]] = S.</cmath> To prove that$ (Error compiling LaTeX. ! Missing $inserted.)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.$$(Error compiling LaTeX. ! Missing inserted.)a = 0, b = 1 \rightarrow S = 1, T = 1.$$ (Error compiling LaTeX. ! Missing$ inserted.)a = 0, b = 2 \rightarrow S = 2, T = 2.$$(Error compiling LaTeX. ! Missing  inserted.)a = 0, b = 3 \rightarrow S = 3, T = 3.$$ (Error compiling LaTeX. ! Missing $inserted.)a = 0, b = 4 \rightarrow S = 4, T = 4.$$(Error compiling LaTeX. ! Missing inserted.)a = 1, b = 1 \rightarrow S = 2, T = 2.$$ (Error compiling LaTeX. ! Missing$ inserted.)a = 1, b = 2 \rightarrow S = 3, T = 3.$$(Error compiling LaTeX. ! Missing  inserted.)a = 1, b = 3 \rightarrow S = 3, T = 4.$$ (Error compiling LaTeX. ! Missing $inserted.)a = 1, b = 4 \rightarrow S = 5, T = 5.$$(Error compiling LaTeX. ! Missing inserted.)a = 2, b = 2 \rightarrow S = 4, T = 4.$$ (Error compiling LaTeX. ! Missing$ inserted.)a = 2, b = 3 \rightarrow S = 4, T = 5.$$(Error compiling LaTeX. ! Missing  inserted.)a = 2, b = 4 \rightarrow S = 5, T = 6.$$ (Error compiling LaTeX. ! Missing $inserted.)a = 3, b = 3 \rightarrow S = 6, T = 6.$$(Error compiling LaTeX. ! Missing inserted.)a = 3, b = 4 \rightarrow S = 6, T = 7.$$ (Error compiling LaTeX. ! Missing$ inserted.)a = 4, b = 4 \rightarrow S = 6, T = 8.$Thus, we have proved for all x and y between 0 and 1, exclusive, <cmath>[5x] + [5y] = T \ge S \ge [3x + y] + [3y + x].</cmath> Now, we prove the lemma statement without restrictions on x and y. Let$ (Error compiling LaTeX. ! Missing $inserted.)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

<cmath>[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}].</cmath>

But, because$(Error compiling LaTeX. ! Missing$ inserted.)[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 <cmath>[5{x}] + [5{y}] \ge [3{x} + {y}] + [3{y} + {x}],</cmath> which we have already proved (as$ (Error compiling LaTeX. ! Missing $inserted.)0 \le {x}, {y} < 1$). 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$(Error compiling LaTeX. ! Missing$ inserted.)p $. We have to prove the exponent of$p $in
$
I = \frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}$</center> is non-negative, or equivalently that <cmath>\sum_{k=1}^{\infty} \left( \left[ \frac{5m}{p^k} \right] + \left[ \frac{5n}{p^k} \right] \right) \ge \sum_{k=1}^{\infty} \left( \left[ \frac{m}{p^k} \right] + \left[ \frac{n}{p^k} \right] + \left[ \frac{3m+n}{p^k} \right] + \left[ \frac{3n+m}{p^k} \right] \right).</cmath> But, the right-hand side minus the left-hand side of this inequality is <cmath>\sum_{k=1}^{\infty} \left( \left[ \frac{5m}{p^k} \right] + \left[ \frac{5n}{p^k} \right] - \left( \left[ \frac{m}{p^k} \right] + \left[ \frac{n}{p^k} \right] + \left[ \frac{3m+n}{p^k} \right] + \left[ \frac{3n+m}{p^k} \right] \right),</cmath> which is the sum of non-negative terms by the Lemma. Thus, the inequality is proved, and so, by considering all primes$ (Error compiling LaTeX. ! Missing $inserted.)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$

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