Three consecutive integers have a sum of 87 .. What are the integers ?!

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Three consecutive integers have a sum of 87 .. What are the integers ?!

Mathematics
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The options are endless...Any answer choices?
No...
Let n-1 be an integer. The next integer is n. The next integer after that would be n+1. so just solve the equation: (n-1)+n+(n+1)=87

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Other answers:

What he said xD Any questions?
I get 2n^3=86 then what?
oh mine how did you get that
Did you tell her to divide 86 into 3 first THEN do this?
how many n's do you see here n, n , n?
3..
n+n+n=3n
-1+1=0
\[(n-1)+n+(n+1)=87 \\(n+n+n)+(-1+1)=87 \\ 3n+0=87 \\ 3n=87\]
n+n+n=3n just like 4+4+4=3(4) or 12
So answer is 29..30..31?
3n=87 gives n=29 n-1=? n+1=?
28,29,30
put n to be 29 these questions I just asked you are equivalent to: n-1=29-1=28 n+1=29+1=30 yes 28+29+30=87
yep we had to find the number before and after 29 since we used (n-1) and n and (n+1) to be are integer where n we found to be 29
Thank you .. Can u stick around and help me with some other problems?
I can probably help with one more
Write 7/3 as a mixed number
How do I start?
how many times does 3 go into 7? or how many 3's are in 7?
2.3
7=3+3+1 how many 3's do you see?
2
right so far we have: \[\frac{7}{3}=2 \frac{?}{3}\]
the question mark is the remainder
1
7=3+3+1 right that 1 right there was left over
so the question mark is 1
Thank you!
np

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