multiples of 400: 400, 800, 1200, 1600, 2000, 2400, ...
That the sum of its digits add up to 9 as for example 9*9 = 81 and 8+1 = 9
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.
Oh, isn't that a happy little math problem! We take the sum of all the odd multiples of five from 1 to 2007, which is 5 + 15 + 25 + ... + 2005. This sum is 100,040,000. When we divide this by ten, we get a remainder of 0, like a peaceful little stream flowing gently through the forest.
There are infinitely many multiples of 9 and it is not possible to add them all.
The sum of consecutive numbers (starting with 1), is the square of the number of terms you sum, in this case 1600. 1600^2 = 2,560,000
Sum of the first 15 positive integers is 15*(15+1)/2 = 120 Sum of the first 15 multiples of 8 is 8*120 = 960
The sum of three consecutive multiples of 6 is 666, the multiples are 216, 222 and 228.
The sum of the first 10 multiples of 3 is 165.
multiples of 400: 400, 800, 1200, 1600, 2000, 2400, ...
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.
That isn't true.
Sum of first 25 multiples of 44+8+12....100taking 4 common4(1+2+3+4....+25) = 4*325 =1300
That the sum of its digits add up to 9 as for example 9*9 = 81 and 8+1 = 9
4*n(n+1)
1 x 6 = 62 x 6 = 12The sum of 12 + 6 = 18
3937