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
The odd multiples of 5 end with 5.If there is an even number of odd multiples of 5, their sum ends in 0, meaning their remainder when divided by 10 is 0;If there is an odd number of odd multiples of 5, their sum ends in 5, meaning their remainder when divided by 10 is 5.By "odd multiples of 5 from 1 to 2007 " do you mean:the multiples 1, 3, 5, ..., 2007 of 5?1 = 2 x 1 - 1, 3= 2 x 2 - 1, ..., 2007 = 2 x 1004 - 1 So there are 1004 multiples of 5, which is even so their sum has a remainder of 0 when divided by 10.the numbers between 1 and 2007 which are odd multiples of 5?1 ÷ 5 = 0.2 → first multiple of 5 in the range is 1 x 5 → the first odd multiple = 1 x 5 2007 ÷ 5 = 401.4 → last multiple of 5 in the range is 401 x 5 → the last odd multiple = 201 x 5401 = 2 x 201 - 1 which means there are 201 multiples of 5, which is odd so their sum has a remainder of 5 when divided by 10.As our number system is based on 10, the remainder when dividing by 10 gives the value of the units (last) digit of the original number.All even multiples of 5 end have a units digit of zero and do not affect the units digit of the sum when added; so the question could have included the even multiples of 5 (ie "What is the remainder when the multiples of five...") without changing the final answer.
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.
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
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