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.
23 x 5=115
The numbers are 21, 22, 23, 24, and 25.
115 + 116 = 131
113, 114, 115
21,22,23,24,25---The Algebra Solution MethodFor any 5 consecutive numbers, starting with X, the four following are X+1, X+2, X+3, and X+4.The equation becomes X + (X+1) + (X+2) + (X+3) + (X+4) = 115So 5x + 10 = 115 and X =21---The Arithmetic Mean MethodAny five consecutive numbers will have an average or mean equal to the middle number. Knowing that 5 times that number is 115,5M = 115 and the central number M is 23.
The numbers are 111, 113 and 115.
114, 115 & 116
23*5 = 115 So pick any four numbers: integers, fractions, irrational numbers, whatever. Add them together to give a sum S. Let the 5th number be 115 - S Then: sum of all five numbers = (sum of the first four numbers) + (the fifth number) = S + (115 - S) = 115 And the count of numbers = 5 So mean = Sum/Count = 115/5 = 23
All numbers ending in five are odd.
Yes. Five is a factor of all cardinal (whole) numbers ending in five
The first ten consecutive composite positive integers are: 114 115 116 117 118 119 120 121 122 123
Chose any integer x. Three consecutive odd numbers: 2x+1 2x+3 2x+5 2x+5+2(2x+1)=115 2x+5+4x+2=115 6x+7=115 6x=108 x=18 The middle number is 2x+3=2*18+3 = 39
57 and 58.