Chat with our AI personalities
They are numbers of the form 2n and 2n+2 where n is any integer.
Let 2n be the first integer, then the next on is 2n+2 and the one after is 2n+4 so adding all three we have 6n+6=108 or 6n=102 and we want 2n so divide we divide 6n by 3 and we have 2n=34, and 2n+2=36 and lastly 2n+4=38 so 34, 36, 38 are the even integers we seek.
150, 152, 154Let the three consecutive even numbers be 2n-2, 2n & 2n+2, then:(2n-2) + 2n + (2n+2) = 456⇒ 6n = 456⇒ n = 76Which means the three numbers are:2n-2 = 2 x 76 - 2 = 1502n = 2 x 76 = 1522n+2 = 2 x 76 + 2 = 154
Two odd numbers added together will always be an even number. I'll show how: Let M and N be any two integers. 2M is even, and 2N is even. So (2M +1) is an odd number, and (2N + 1) is an odd number. (2M +1) + (2N +1) = 2M + 1 + 2N +1 = 2M + 2N + 2 = 2(M + N + 1), which is even.
Answer116, 118, 120, 122, 124Procedureassume the numbers are such that the middle one is 2n,the assumed numbers are: 2n-4, 2n-2, 2n, 2n+2, 2n+4 ..... (1)Summation = 10n = 600so, n = 60Substitute with n = 60 in (1) we get the answer above