Q: What is the sum of three even consecutive integers is 198?

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64 + 66 + 68 = 198

They are: 65+66+67 = 198

198 is an even number.

Let the first even number be x. Then the second even number would be x + 2, and the third even number would be x + 4. The sum of these numbers is x + (x + 2) + (x + 4) = 3x + 6. We set this equal to 204 and solve for x: 3x + 6 = 204 --> 3x = 198 --> x = 66. Therefore, the three consecutive even numbers are 66, 68, and 70.

Let n be an integer, then two consecutive odd integers can be expressed as n and n+2. The equation can the be wrote as n2 + (n+2)2 = 202 = n2 +n2 + 4n +4 = 202 which implies 2n2 + 4n - 198 = 0 which implies n2 + 2n - 99 = 0 the quadratic formula can then be used to solve for n as n = (-2 +- (4 - 4*-99)^(1/2))/(2) = -1 +- (1/2)*(400)^(1/2) = - 1 +- 10 take the positive solution and you get n = 9 which implies your two consecutive integers are 9 and 11.

Related questions

201

The numbers are 65, 66, and 67..

64 + 66 + 68 = 198

48, 49, 50, 51

They are: 65+66+67 = 198

198, 199 and 200 if you add them.

198 is an even number.

Let the first even number be x. Then the second even number would be x + 2, and the third even number would be x + 4. The sum of these numbers is x + (x + 2) + (x + 4) = 3x + 6. We set this equal to 204 and solve for x: 3x + 6 = 204 --> 3x = 198 --> x = 66. Therefore, the three consecutive even numbers are 66, 68, and 70.

198 is the greatest three-digit even number that has no factor equal to 4.

Let n be an integer, then two consecutive odd integers can be expressed as n and n+2. The equation can the be wrote as n2 + (n+2)2 = 202 = n2 +n2 + 4n +4 = 202 which implies 2n2 + 4n - 198 = 0 which implies n2 + 2n - 99 = 0 the quadratic formula can then be used to solve for n as n = (-2 +- (4 - 4*-99)^(1/2))/(2) = -1 +- (1/2)*(400)^(1/2) = - 1 +- 10 take the positive solution and you get n = 9 which implies your two consecutive integers are 9 and 11.

The numbers are 195, 196, 197 and 198.

198 is composite because it has three or more factors.