Question:

The HCF of two consecutive even numbers is:

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Consecutive integers are co-prime. For consecutive even numbers \(2n\) and \(2n+2\), a factor \(2\) is always common, so HCF \(=2\).
Updated On: Oct 27, 2025
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The Correct Option is C

Solution and Explanation

Step 1: Represent consecutive even numbers.
Let the two consecutive even numbers be \(2n\) and \(2n+2 = 2(n+1)\).
Step 2: Take out the common factor.
\(\gcd(2n,\,2n+2)=\gcd\big(2n,\,2(n+1)\big)=2\cdot \gcd\big(n,\,n+1\big).\)
Step 3: Use property of consecutive integers.
\(\gcd(n,\,n+1)=1\) (consecutive integers are co-prime). Hence, the HCF is \(2\times 1=2\).
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