Show that any positive even integer …

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Sia ? 6 years, 4 months ago
Let p be any positive integer
By division algorithm, p = 6m + r, where 0 {tex} \leqslant {/tex}r< 6
Here r=0,1,2,3,4,5
Therefore,values of p are : 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5
Now 6m+1,6m+3 and 6m+5 are odd numbers because m is a positive integer.
Hence 6m, 6m + 2, 6m + 4 are even integers because they are next positive number to the odd numbers 6m-1,6m+1 and 6m+3 respectively
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