Help me !

My math teacher gave me a lot of problems to solve. But I don't have to worry because these problems aren't so difficult to solve. But one thing made me worried. That's why I need a little help. Tell me how can I prove that the product of any two consecutive even numbers is divisible by 8.

Please......help me! Quick!

#ProofTechniques

Easy Math Editor

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Comments

It's quite simple to think. Choose any two consecutive even no and u wil see only one of them is divisible by 4 but other is not. Now u r dividing the product of this two by 8 witch can be written as 4*2. Now look at the nos u have chosen before , one of them will be divisible by 4 and other will be divisible by 2.. nw u r done.. the product of ur chosen no are divisible by 8.

Sayan Kumar - 7 years, 4 months ago

I'll help you (Please follow me in return!)

See, every consecutive pair of even numbers can be expressed as 2x and 2x+2 or 2(x+1). Their product will be 2 * 2 * x * (x+1)=4x(x+1). Now, if x is even, then 4x(x+1) will be divisible by 8. And if x is odd, then (x+1) will be even. So 'the product of any two consecutive even numbers is divisible by 8'. Please follow me for excellent problems on number theory and algebra.

Satvik Golechha - 7 years, 4 months ago

First you follow me, then I will follow you.

Raiyun Razeen - 7 years, 4 months ago

let one of them is $2n$ . then other one is $2n+2$ . now $\left( 2n+2 \right) *2n=4*(n+1)*n$

now $n$ and $n+1$ are two consecutive number. therefore one of the is even and other one is odd therefore $n*(n+1)$ must be divisible by 2

therefor $n*\left( n+1 \right) =2*k$ where $k$ is an integer

therefore $\left( 2n+2 \right) *2n=4*2*k$

Bedadipta Bain - 7 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$