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Sagot :
Every even number is 2 away from the last.
0, 2, 4, 6, 8, 10, 12, 14...etc.
If we had an even number p, then the next three even numbers would be
p+2, p+4, and p+6.
(If we had an odd number p, then the next three even numbers would be
p+1, p+3, and p+5. I'm not sure if p is even is implied in the question. Technically the answer would be p - p mod 2 + 2, where p is an interger...that gets into more technical function stuff, though.)
0, 2, 4, 6, 8, 10, 12, 14...etc.
If we had an even number p, then the next three even numbers would be
p+2, p+4, and p+6.
(If we had an odd number p, then the next three even numbers would be
p+1, p+3, and p+5. I'm not sure if p is even is implied in the question. Technically the answer would be p - p mod 2 + 2, where p is an interger...that gets into more technical function stuff, though.)
We don't know whether ' P ' itself is odd or even.
-- If ' P ' is even, then the next three [larger] even numbers are
P+2, P+4, and P+6 .
-- If ' P ' is odd, then the net three [larger] even numbers are
P+1, P+3, and P+5.
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