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There are 4 consecutive even integers that add up to 100. What is the least of the 4 integers?

User Jassuncao
by
6.7k points

1 Answer

3 votes

Answer:

22

Explanation:

Let's say the 4 consecutive integers are a, b, c, and d

a=b-2 ==> Consecutive even numbers have a difference of 2: Ex. (4-2=2)

b=c-2

c=d-2

a+b+c+d=100

Solve for a:

a=b-2 ==> add 2 on both sides

a+2=b

a+(a+2)+c+d=100 ==> substitute a+2 for b

a+2=c-2 ==> b=c-2 and a+2=b, so a+2=c-2

a+2+2=c-2+2 ==> add 2 on both sides

a+4=c

a+(a+2)+(a+4)+d=100 ==> substitute a+4 for c

a+4=d-2 ==> c=d-2 and a+4=c, so a+4=d-2

a+4+2=d-2+2 ==> add 2 on both sides

a+6=d

a+(a+2)+(a+4)+(a+6)=100 ==> substitute a+6 for d

a+a+2+a+4+a+6=100 ==> solve for a

a+a+a+a+2+4+6=100

4a+12=100 ==> simplify

4a+12-12=100-12 ==> isolate a

4a=88 ==> simplify

a=88/4

a=22

User Jkigel
by
6.6k points
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