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Three consecutive odd integers that have a sum of 123?
123/3 = 41 so integers are 39, 41 and 43
Ind three consecutive odd integers that have a sum of 123?
x is lowest of the integers so x + (x + 2) + (x + 4) = 123 ie 3x + 6 = 123 so 3x = 117 and x = 39, so integers are 39, 41 and 43
What is the sum of three consecutive integers is 126 so what are the integers?
:/ We can write it this way: x - 1 + x + x + 1 = 126. 1s cancel out, so we get 3x = 126. 126/3 = 42. Answer: 41, 42, 43.
What are three consecutive odd integers that have a sum of 123?
39, 41, 43 Let x represent the smallest of these numbers. From the problem, we know x+(x+2)+(x+4)=123 Solving for x: 3x+6=123 3x=117 x=117/3=39 So our integers are: 39, 41, 43
How would you reconise a multiple of three?
if the number is a 2 or more diget number, you add up all of the numbers in the number and when you have the sum, if the sum is divisible by 3 , the whole number is divisible by three EX. 123=6, 6 is divisible by 3 so 123 is divisible by three 447=15 is divisible by 3 so 447 is divisible by 3 989=26 is *NOT* divisible by 3 so 989 is not divisible by three this truly does work for every single number
