find three consecutive integers such that the sum of the first and the third is 63?

Answer 1

x +x+2=63
=> x = 30.5

consecutive integers : impossible

Answer 2

This is an impossible situation.

Let the integers be a, b and c.

Since they are consecutive, that means that a and c will either both be odd or both be even. In either case, adding and an odd to an odd or an even to an even will give you an even answer. Therefore, the number 63 cannot be the sum of two integers in this case.

However, I will work out how to solve for a and c anyway to prove that they are not integers.

a + 1 = b
b + 1 = c or a + 2 = c

a + c = 63
a + a + 2 = 63
2a + 2 = 63
2a = 61
a = 30.5

30.5 is not an integer. Therefore, the problem has no solution.

Answer 3

It's not possible.
Let x = the middle number
then x-1 is the first number
and x+1 is the third number
(x-1) + (x+1) = 63 is what you're asking
2x = 63
x = 31.5 (not an integer)

If the problem is to find three consecutive intergers that add to 63, then it's 20, 21, 22.

If the problem is to find three consecutive ODD intergers that add to 63, then it's 19, 21, 23.

The sum of the first and third numbers of three consecutives integers will always be even.

Answer 4

No solution.

30+32=62
31+33=64

I think you must mean: Three consecutive integers such that the PRODUCT of the first and the third is 63. The answer to that is 7, 8, 9.

Answer 5

This is an impossible situation. 30.5,31.5,32.5 would work, but they are not integers. I tried negative numbers but it did not work, so I am sorry to say that if that is a homework problem you will not get any points for it.

Answer 6

x, x+1, x+2

x+x+2=63
2x=61
x=30.5
not an integer, therefore it cannot be done

Answer 7

HI!!! is right. No integers would make this work. 30.5, 31.5, 32.5 would work, but they are not integers.

Answer 8

impossible!