Question

Prove that the product of any 'm'consecutive integers is divisible by 'm'

Answer 1

Step by step explanation :

1. First of all m consecutive integers are taken .

2. Then Applying Common algorithm on "c" and "m" , it is proved that "m" divides "c" for all values of c ..

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1. প্ৰথমে একেৰাহে পূৰ্ণসংখ্যা লোৱা হয়।

2. তাৰ পিছত "গ" আৰু "এম"ত উমৈহতীয়া এলগৰিথম প্ৰয়োগ কৰিলে, এইটো প্ৰমাণিত হয় যে "এম"-এ গ.-ৰ সকলো মূল্যৰ বাবে "গ" বিভাজন কৰে।

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Answer 2

The product of any 'm' consecutive integers is divisible by 'm'.

Concept of Division Algorithm:

The conventional long division algorithm consists of a set of steps that are repeated in the following order: divide, multiply, subtract, and bring down.

Given:

Product of 'm' consecutive numbers are to be taken.

Explanation:

Let  $m$ consecutive integers may be taken as $c, c+1, c+2, c+3, \ldots \ldots$$c+m-1$ where $c$ is any integer.

Applying Division Algorithm on $c$ and $m$ we have,

$\begin{aligned}c &=m q+r \quad 0 \leq r < m \\ \quad & c=m q, m q+1, m q+2, \ldots \ldots, m q+m-1\end{aligned}$

Now, for $c=q m, m \mid c$

For $c=q m+1, c+(m-1)=q m+1+m-1=(q+1) m$

i.e., $m \mid c+(m-1)$

Similarly for, $c=q m+2, c+(m-2)=q m+2+(m-2)=(q+1) m$

i.e., $m \mid c+(m-2)$

And for, $c=q m+m-1$, $c+1=q m+m-1+1=(q+1) m$ i.e., $m \mid c+1$

Therefore, for any integral value of $c$, one of $c, c+1, c+2, \ldots ., c+(m-1)$ divisible by $m .$

Thus, $m \mid c(c+1)(c+2) \ldots \ldots .(c+m-1)$ is divisible by $m$.

To know more about Division Algorithm, here

https://brainly.in/question/1618095

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