Q: What are three consecutive multiples of 17?

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20 + 25 + 30 = 75

Let the second of the three consecutive multiples of 6 be 6n Then the first is 6n - 6 and the last is 6n + 6; and: (6n - 6) + 6n + (6n + 6) = 666 → 18n = 666 → n = 37 → the consecutive multiples of 6 which sum to 666 are 216, 222, 228

The only proper factor of 51 that is less than 1/3 of 51 is 3. Divide the sum of three consecutive odd integers by 3: 51/3 = 17. The smallest of these integers will be two less than 17 and the largest will be two more than 17, so the three consecutive odd integers having a sum of 51 are 15, 17, and 19.

All multiples of 17 are divisible by 17. There is an infinite number of multiples of 17.

Consecutive numbers are those in a continuous sequence for a given subset of numbers, usually integers. Example of consecutive integers: 6, 7, 8, 9 Consecutive odd numbers: 7, 9, 11, 13 Consecutive even numbers: 6, 8, 10, 12 Consecutive multiples of 10: 30, 40, 50, 60 Consecutive prime numbers: 5, 7, 11, 13, 17

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The sum of three consecutive multiples of 6 is 666, the multiples are 216, 222 and 228.

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If half of the sum is 90 then the full sum must be 180. The three consecutive multiples of 10 which add up to 180 are 50, 60 and 70.

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20 + 25 + 30 = 75

45, 60, and 75

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45, 60, and 75

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First find the number which is 1/3 of 51. This is 17. Thus any three numbers which average 17 will always add up to 51. For the numbers to be consecutive and have an average of 17, the middle number must be 17 and the two remaining numbers are 17-1 and 17+1 ie 16 and 18. So the three consecutive numbers are 16, 17 and 18.