No. There is no set of three consecutive even integers with a sum of 40.
31 32 33
The integers are -37, -36 and -35. Also, using consecutive even integers: -38, -36 and -34.
They are -12, -10 and -8.
There are no two consecutive integers whose product is 421 - the product of 20 and 21 is 420.
117/3 = 39, so the three consective odd integers whose sum is 117 are 37, 39, and 41.
63/2 = 31.5 so the two are 31 and 32.
16 17 18 How to get the answer.... 51/3 = 17 17+1= 18 17-1= 16
23+21+19+17=80
the sum of three consective even integers is negative 300
63, 65, 67, 69 = 264 so there does not appear to be a solution in whole numbers!
I'm going to assume that you mean three not tree. Below are the steps to follow to answer this question. # divide 36 by 3 giving you 12 # 12 + 12 + 12 = 36 # subtract 2 from the first 12 and add it the last one # (12-2) + 12 + (12+2) = 36 # 10 + 12 + 14 = 36 # three consective even integers are 10, 12, and 14
The three integers are -27, -26 and -25.
The integers are -1, 1 and 3.
There is no set of three consecutive integers whose sum is 71.
Find 2 consecutive ODD integers whose sum is -88 ANSWER: -45, -43
The integers are -16 and -14.