Let p be a prime and m a positive integer. By mathematical induction on m, or otherwise, prove that

Q: Why can't we just expand the factorials? It looks like a nightmare.

Expanding factorials is like trying to solve a maze by walking into every wall. It is possible, but it is painful and prone to error. The identity $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$ is our 'shortcut'—it respects the structure of the binomial coefficient rather than destroying it. By using the identity, we keep the expression clean and manageable, allowing us to isolate the prime $p$ without getting lost in a sea of factorials.

Q: How do we know $\frac{m}{r} \binom{mp-1}{r-1}$ is an integer?

This is the 'Integrality Argument.' We know $\binom{mp}{r}$ is an integer because it represents the number of ways to choose $r$ items from $mp$ items. If $\binom{mp}{r} = p \cdot X$, and we prove $X$ must be an integer, we are done. Since $r$ divides the numerator $p \cdot m \cdot \binom{mp-1}{r-1}$ and $\gcd(r, p)=1$, Euclid's Lemma guarantees $r$ divides the rest. Therefore, $X$ must be an integer.

Q: Does this proof work if $p$ divides $r$?

Great catch! No, it does not. The condition $p \nmid r$ is absolutely critical. If $p$ were to divide $r$, the prime factor $p$ in the numerator could be 'cancelled out' by the $p$ hidden inside $r$, and the divisibility property would fail. The beauty of this problem lies entirely in the fact that $p$ and $r$ are coprime.

Question

Let $p$ be a prime and $m$ a positive integer. By mathematical induction on $m$, or otherwise, prove that whenever $r$ is an integer such that $p$ does not divide $r$, $p$ divides $^{mp}C_r$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey into the heart of number theory and combinatorics. We are going to prove a beautiful property: that for any prime $p$ and positive integer $m$, if $p$ does not divide $r$, then $p$ must divide the binomial coefficient $\binom{mp}{r}$.

When you look at $\binom{mp}{r}$, your instinct might be to write out the factorials:

$$\frac{(mp)!}{r!(mp-r)!}$$

Stop. Take a breath. That path leads to a forest of terms that will obscure the truth.

### The Power of the Identity

Instead, we use the most powerful tool in our combinatorial arsenal: the identity $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$. This is powerful because it allows us to extract the variables $n$ and $k$ from the binomial coefficient without destroying the structure of the remaining term.

By substituting $n = mp$ and $k = r$, we transform our expression into:

$$\binom{mp}{r} = \frac{mp}{r} \binom{mp-1}{r-1}$$

We have successfully pulled the $mp$ out.

### Isolating the Prime

Now, we want to show that $p$ divides this entire expression. Let us isolate $p$ by rewriting our equation as:

$$\binom{mp}{r} = p \cdot \left( \frac{m}{r} \binom{mp-1}{r-1} ight)$$

To prove that $p$ divides $\binom{mp}{r}$, we simply need to prove that the term inside the parentheses, $\frac{m}{r} \binom{mp-1}{r-1}$, is an integer. If it is an integer, then $\binom{mp}{r}$ is clearly a multiple of $p$.

### The Logic of Integrality

This is where the physical reality of the problem saves us. $\binom{mp}{r}$ represents the number of ways to choose $r$ objects from a set of $mp$ objects, which by definition must be an integer.

Therefore, the right-hand side, $p \cdot \left( \frac{m}{r} \binom{mp-1}{r-1} ight)$, must also be an integer. This implies that the denominator $r$ must perfectly divide the numerator $p \cdot m \cdot \binom{mp-1}{r-1}$.

### The Final Piece: Euclid's Lemma

We are given that $p$ is a prime number and $p 
mid r$. This means that $p$ and $r$ are coprime, or $\gcd(p, r) = 1$.

Since $r$ has no common factors with $p$, it cannot 'cancel out' the $p$ in the numerator. By Euclid's Lemma, if $r$ divides a product $A \cdot B$ and $\gcd(r, A) = 1$, then $r$ must divide $B$.

Here, $r$ must divide $m \cdot \binom{mp-1}{r-1}$. This means that the fraction $\frac{m}{r} \binom{mp-1}{r-1}$ is, in fact, an integer. Let us call this integer $K$.

We have arrived at the beautiful conclusion:

$$\binom{mp}{r} = p \cdot K$$

By the very definition of divisibility, this proves that **$p$ divides $\binom{mp}{r}$**. You have just navigated through a classic proof that bridges the gap between simple counting and deep number theory.

Animated Solution for Mathematics - Binomial Theorem: Let $p$ be a prime and $m$ a positive integer. By mathematical induction on $m$ , or otherwise, prove that whenever $r$ is an integer such that $p$ does not divide $r$ , $p$ divides $^{m p} C_{r}$ .

Visualized Solution

The Problem Statement

The Combinatorial Identity

Applying the Identity

Isolating the Prime $p$

The Property of Integrality

The Role of the Denominator

Coprimality of $p$ and $r$

Applying Euclid's Lemma

Defining the Integer $K$

Final Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Power of the Identity

Isolating the Prime

The Logic of Integrality

The Final Piece

Similar Questions

Using mathematical induction, prove that $^{m} C_{0}^{n} C_{k} +^{m} C_{1}^{n} C_{k - 1} + .... +^{m} C_{k}^{n} C_{0} =^{(m + n)} C_{k}$ , where $m, n, k$ are positive integers, and $^{p} C_{q} = 0$ for $p < q$ .

Using induction or otherwise, prove that for any non-negative integers $m, n, r$ and $k$ , $\sum_{m = 0}^{k} (n - m) \frac{( r + m )!}{m !} = \frac{( r + k + 1 )!}{k !} [\frac{n}{r + 1} - \frac{k}{r + 2}]$

For any positive integer $m, n$ (with $n \geq m$ ), let $(m n) =^{n} C_{m}$ . Prove that $(m n) + (m n - 1) + (m n - 2) + .... + (m m) = (m + 1 n + 1)$ . Hence or otherwise, prove that $(m n) + 2 (m n - 1) + 3 (m n - 2) + .... + (n - m + 1) (m m) = (m + 2 n + 2)$ .

Prove that $C_{0} - 2^{2} C_{1} + 3^{2} C_{2} - .... + (- 1)^{n} (n + 1)^{2} C_{n} = 0, n > 2$ , where $C_{r} =^{n} C_{r}$ .

Prove that $\sum_{r=1}^k (-3)^{r-1} ^{3n}C_{2r-1} = 0$ , where $k = (3 n) /2$ and $n$ is an even positive integer.

Let $n$ be any positive integer. Prove that $\sum_{k = 0}^{m} \frac{( k 2 n - k )}{( n 2 n - k )} (2 n - 4 k + 1) 2^{n - 2 k} = \frac{( m n )}{( n - m 2 n - 2 m )} 2^{n - 2 m}$ for each non-negative integer $m \leq n$ .

Prove that $2^{k} (0 n) (k n) - 2^{k - 1} (1 n) (k - 1 n - 1) + 2^{k - 2} (2 n) (k - 2 n - 2) - .... + (- 1)^{k} (k n) (0 n - k) = (k n)$

Suppose $det [\sum_{k = 0}^{n} k \sum_{k = 0}^{n}^{n} C_{k} k \sum_{k = 0}^{n}^{n} C_{k} k^{2} \sum_{k = 0}^{n}^{n} C_{k} 3^{k}] = 0$ , holds for some positive integer $n$ . Then $\sum_{k = 0}^{n} \frac{^{n} C _{k}}{k + 1}$ equals

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

Given $s_{n} = 1 + q + q^{2} + .... + q^{n}$ ; $S_{n} = 1 + \frac{q + 1}{2} + (\frac{q + 1}{2})^{2} + .... + (\frac{q + 1}{2})^{n}, q \neq = 1$ Prove that $^{n + 1} C_{1} +^{n + 1} C_{2} s_{1} +^{n + 1} C_{3} s_{2} + .... +^{n + 1} C_{n + 1} s_{n} = 2^{n} S_{n}$

Visualized Solution

The Problem Statement

The Combinatorial Identity

Applying the Identity

Isolating the Prime .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

The Property of Integrality

The Role of the Denominator

Coprimality of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Applying Euclid's Lemma

Defining the Integer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Final Conclusion

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Power of the Identity

Isolating the Prime

The Logic of Integrality

The Final Piece

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Using induction or otherwise, prove that for any non-negative integers m,n,r and k, ∑m=0k​(n−m)m!(r+m)!​=k!(r+k+1)!​[r+1n​−r+2k​]

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