Question

Find all pairs of consecutive even positive integers both of which are larger than 8, such that their sum is less than 25.

Answer 1

Step-by-step explanation:

Let the smaller and even the positive integer be x,

Since the larger integer is consecutive even,

It will be x+2

Given that

both the integers are larger than 5,

that is x > 5

and x +2 >5

x>5-2

x>3

Since x>3 and x>5

x>5

Also,given that

Sum of the two integers is less than 23.

Therefore,x+(x+2)<23

2x +2 <23

2x <23 -2

2x <21

x< 21/2

x <10.5

Hence, x>5 and x<10.5

Integers greater than 5 but less than 10.5 are 6,7,8,9,10

Even integers greater than 5 but less than 10.5 are 6,8,10

hence x=6,8,10.

      x                                  x+2                                     Pair (x,x+2)

• 6                                 6 +2=8                               6,8

• 8                                 8+2=10                               8,10

• 10                               10+2=12                             10,12

Thus,the required possible pairs are (6,8), (8,10), (10,12)

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