There are 67 multiples of 6 and 50 multiples of 8 in that range. Their total, 117, will include numbers that are both.
The first six multiples of 3 are 3, 6, 9, 12, 15, and 18.
66 is the first one
3, 6, 9, 12, 15, 18 4, 8, 12, 16, 20, 24 6, 12, 18, 24, 30, 36 Common multiples include 12, 24, 36, 48, 60, 72
2, 4, 6, 8, 10, 12
The sum of all multiples of 3 below 500 is the sum of 3 + 6 + ... + 498 = 41583The sum of all multiples of 5 below 500 is the sum of 5 + 10 + ... + 495 = 24750Depending upon the interpretation of "of 3 and 5", the answer is one of:The sum of all multiples of both 3 and 5 below 500 is the sum of 15 + 30 + ... + 495 = 8415The sum of all multiples of 3 below 500 and all multiples of 5 below 500 is 41583 + 24750 = 66333Since the multiples of both 3 and 5 (that is 15, 30, ...) have been counted twice - once in the multiples of 3 and once in the multiples of 5 - the sum of all multiples of 3 or 5 or both below 500 is 41583 + 24750 - 8415 = 57918I have emphasised the word below with regard to 500 since 500 is a multiple of 5, but 500 is not below (less than) itself, that is the multiples are of the numbers 1, 2, 3, ..., 499. If the question was intended to mean less than or equal to 500, add an extra 500 to the multiples of 5 above, making the sum of the multiples of 5 be 25250 and the final sums (1) 8415, (2) 66833, (3) 58418.
90
5 + 10+15+20=50 ans
It is: 6+12+18 = 36
The sum of three consecutive multiples of 6 is 666, the multiples are 216, 222 and 228.
1 x 6 = 62 x 6 = 12The sum of 12 + 6 = 18
6+12+18+24+30=90
6 + 12 + 18 + 24 + 30 + 36 = 126
The first 6 multiples of 9 are 9, 18, 27, 36, 45 and 54. Their sum (total) is 189. In other words, 1x9 + 2x9 + 3x9 + 4x9 + 5x9 + 6x9 = 21x9 = 189.
Why, that's just 6 times the sum of the numbers from 1 to 50 !The sum of the numbers from 1 to 50 is:1 + 50 = 512 + 49 = 513 + 48 = 514 + 47 = 515 + 46 = 51...etc. There are 25 pairs, and each pair sums to 51. So the sum of 1 ... 50 is (25 x 51) = 1,275.The sum of the first 50 multiples of 6 is (1,275 x 6) = 7,650.
There are 83 multiples of six that fall between 500 and 1,000.
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