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Let the smallest of the three consecutive integers be 'x'.

Then the three numbers can be represented by x, (x + 1) and (x + 2)

Then, 2(x + 2) - 15 = x (twice the third is 15 more that the first)

2x + 4 - 15 = x

2x - x = 15 - 4

x = 11

The three numbers are thus : 11, 12, and 13.

And, 2 x 13 = 26 is 15 more than the first number 11.

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Q: How would you find 3 consecutive integers such that twice the third is 15 more than the first?
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