Let the consecutive numbers be n,n+1, n+2, n+3, & n+ 4.
Add
n + (n+1)+(n+2) + (n + 3) + ( n + 4) = 115
5n + 10 = 115
5n = 105
n = 21
N+1 = 22
n+2 = 23
n+3 = 24
n + 4 = 25
So the five consecutive numbers are 21,22,23,24 & 25.
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You have five numbers that add up to 115. What is the average value of those five numbers? Now, how do you choose five consecutive numbers that have that average value?
The numbers are 21, 22, 23, 24, and 25.
Let 'n' be the first odd number,. Then n+ 2 & n + 4 are the second and third odd number. Hence n + (n+2) + ( n + 4) = 339 Add 3n + 6 = 339 3n = 333 n = 111 n + 2 = 113 n + 4 = 115 Hence the three odd consecutive numbers are 111,113 & 115.
Since 115 ends in five, it will divide by '5'. Hence '5' is one of the factors. 5)115 = 23 Hence '23' is the other factor. Since '23' is a prime number, there are no other factors for 115 , but 5, and 23. except '1' & 115'.
gcf for these numbers 345 253 and 115 is 23.
115 is one hundred and fifteen.