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!

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TopNewestIt'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 · 2 years, 11 months ago

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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 · 2 years, 11 months ago

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– Raiyun Razeen · 2 years, 11 months ago

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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 · 2 years, 11 months ago

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