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Exercise 5.1 · Q24

Q.Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

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Let the first even integer be xx; the next is x+2x+2. With x>5x > 5 and x+(x+2)<23x + (x+2) < 23, the valid pairs are (6,8)(6,8), (8,10)(8,10), and (10,12)(10,12).

Setting up the conditions. Let the smaller consecutive even integer be xx, so the larger is x+2x + 2. The problem requires:

  • both larger than 55: since x>5x > 5 automatically makes x+2>5x + 2 > 5, the binding condition is x>5x > 5;
  • their sum less than 2323: x+(x+2)<23x + (x+2) < 23.

Solving the sum inequality.

x+(x+2)<23x + (x+2) < 23

2x+2<232x + 2 < 23

2x<212x < 21

x<10.5x < 10.5

Combining the conditions.

5<x<10.55 < x < 10.5

Since xx must be an even integer, the possible values are x=6, 8, 10x = 6,\ 8,\ 10, giving: …

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