5 + 10+15+20=50 ans
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.
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The first 6 multiples of 5 are: 5, 10, 15, 20, 25, and 30.
75
90
5 + 10+15+20=50 ans
The first seven of them are.
50 By plus i assume you mean sum. The first four multiples of 5 are 5,10,15, and 20 5+10+15+20=50
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.
.....5 .....Σ 9r ...r=1
25,20,5,30
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.
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They are the first three multiples of LCM(5, 9), that is, the first three multiples of 45.
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