Question 254489: The sum of Three consecutive even integers is 196. Find the largest of these integers? Found 3 solutions by drk, solver91311, MRperkins:Answer by drk(1908) (Show Source):
You can put this solution on YOUR website! Let the three consecutive even numbers be
X
X+2
X+4
The sum is 196 translates to
X + X + 2 + X + 4 = 196
3X + 6 = 196
3X = 190
However, 3 doesn't divide into 190 evenly.
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If the sum were 198 instead, we get
3X + 6 = 198
3X = 192
X = 64
X+2 = 66
X+4 = 68
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The answer could also be no solution.
Let represent the largest of 3 consecutive even integers. The next lower consecutive even integer is then , and the next lower one is . Add 'em up to get 196:
Just solve for
The problem is that the value of that satisfies the equation is not an integer, much less an even integer. Therefore, there is no set of three consecutive even integers such that their sum is 196.
As further proof, consider:
Note that the sum is smaller than 196, and then next larger sum of three consecutive even integers:
has a sum larger than 196.
Hence the statement: "The sum of three consecutive even integers is 196." is false and the largest of three non-existent integers is also non-existent.
You can put this solution on YOUR website! even integers are represented by 2n, 2n+2, 2n+4...
We want 3 consecutive even integers to =196, so
1 even integer is 2n
2nd consecutive even integer is 2n+2
3rd consecutive even integer is 2n+4
so: 2n+2n+2+2n+4=196
simplify and get 6n+6=196
6n=190
n=190/6
Possibly, there are not 3 even integers whose sum is equal to 196 (fractions are not integers)
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