three consecutive numbers that add up to 187?

In your update, you say *two* consecutive integers.

Let n be the first number.

Let n+1 be the second number.

The sum is 187:

n + n+1 = 187

2n + 1 = 187

2n = 186

n = 186/2

n = 93

Answer:

93 and 94

1 8 and 7

n + (n + 1) = 187

2n + 1 = 187

2n = 186

n = 93

The numbers are 93 and 94..

187 is NOT divisible by 3. Therefore there exists no triplet of consecutive integers such that the sum is not

divisible by 3. Easy to prove: Put triplet = n-1, n, n+1. Then sum = 3n which IS divisible by 3.

(x-1)+x+(x+1)=187

3x=187

x=62.33...

so the numbers are 61.33..., 62.33..., 63.33... I wouldn't call them consecutive though, they're not integers.

(Bear in mind that 187 isn't divisible by 3 because the sum of 187's digits is 1+8+7=16 which isn't divisible by 3).

I don't know the answer but I know that 61+62+63=186. Is this problem even possible?