Question 102137: FIND FOUR CONSECUTIVE INTEGERS SUCH THAT THE SUM OF THE SECOND AND FOURTH IS 132
Found 2 solutions by checkley75, Kalmetam:
Answer by checkley75(3666)   (Show Source):
You can put this solution on YOUR website! X, X+1, X+2, X+3
X+1+X+3=132
2X+4=132
2X=132-4
2X=128
X=128/2
X=64
X+1=64+1=65 ANSWER FOR THE SMALLER NUMBER.
X+3=64+3=67 ANSWER FOR THE LARGER NUMBER.
PROOF
65+67=132
132=132
Answer by Kalmetam(43) (Show Source):
You can put this solution on YOUR website! Well, half of it must be one of the numbers.. The 2nd and fourth MUST be the ones 1 before and 1 after the half number because 1 over plus 1 under equals the whole number, in this case, 132.. So that must mean that the 3rd number is half! Half is 66. So that is the 3rd number so the rest is obvious..
64,65,66,67
65+67=132
Now we know the answer
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