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Here's how you can work it out:

First, let "x" be the first of those three numbers. We know that those numbers are three consecutive even integers, which tells us that the other numbers are "x + 2" and "x + 4". We also know that the three of them added together give us 210, so we can say:

x + (x + 2) + (x + 4) = 210

∴ 3x + 6 = 210

∴ 3x = 204

∴ x = 68

We now know that the lowest number is 68, which tells us that the other two must be 70 and 72. To check that, we simply add them together:

68 + 70 + 72 = 210

which confirms our answer.

Q: What are 3 consecutive even integers whose sum is 210?

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You cannot.If you have any two consecutive numbers, one of them must be odd and the other even. So their sum must be odd and therefore cannot be 702 nor 210

Let x = first positive integer the x + 1 is the next positive integer the product is x (x+1) = 210 x^2 + x = 210 x^2 + x - 210 = 0 Use quadratic equation to solve, or factor (X+15) (X-14 ) = 0 x = +14 or X = -15 The positve one is x = 14 X + 1 = 15 ANS: 14 and 15

1x2x3=6 2x3x4=24 3x4x5=60 4x5x6=120 5x6x7=210 6x7x8=336 7x8x9=504 8x9x10=720 9x10x11=990

The numbers are 51, 52, 53 and 54.

14 and 15 14 X 15 = 210 7 X 30 = 210

Related questions

Middle integer must be one-third of the sum, in this case 70. Other two are therefore 68 and 72

You can solve this in two ways.1) Trial and error. That is, try multiplying two consecutive integers; if the product is too large, try smaller integers; if the product is too small, try larger consecutive integers. 2) Call the two consecutive integers "n" and "n+1", and solve the equation: n(n+1)=210

210, 211, 212, 213 and 214.

66, 68, 70 Check: 66+68=134, 134+76=210, 210/3=70 ; 66, 68, 70

There are an infinite number of prime numbers which are consecutive odd integers. Choose any natural number n. Take all primes up to any number n, take their product, and add 1 and subtract 1 from it. These 2 numbers are consecutive odd integers. eg 2*3*5*7 = 210 209 and 211 are primes which are consecutive odd integers.

Let's assume that the three consecutive integers are n-1, n, and n+1. We know that their product is 210, so we can set up the equation (n-1)(n)(n+1) = 210. By solving this equation, we can find that the three integers are 6, 7, and 8. Therefore, the sum of the smallest two integers is 6 + 7 = 13.

You cannot.If you have any two consecutive numbers, one of them must be odd and the other even. So their sum must be odd and therefore cannot be 702 nor 210

The Answer:This, we can notate as: (x - 1)x(x + 1) = (x2 - 1)x = x3 - x = 210.x3 - x - 210 = (x - 6)(x2 + 6x + 35) = 0Thus, 6 = x (or -3 +- i*sqrt(26)). The real solution, 6, can be plugged back into x - 1 (the lowest), producing 5. 5 is your answer.

5*6*7 = 210

14 x 15 = 210

Let x = first positive integer the x + 1 is the next positive integer the product is x (x+1) = 210 x^2 + x = 210 x^2 + x - 210 = 0 Use quadratic equation to solve, or factor (X+15) (X-14 ) = 0 x = +14 or X = -15 The positve one is x = 14 X + 1 = 15 ANS: 14 and 15

Consecutive numbers can't both be multiples of 7. The LCM of consecutive numbers is their product. 14 and 15 are consecutive numbers whose LCM is a multiple of 7 that is greater than 200.