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Q: What is the sum of the first 500 multiples of 3?

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975

10+20+30 = 60

The first 50 multiples of 6 are the first fifty even multiples of 3.

1 + 3 + 5 + .... + 995 + 997 + 999 = 250000

All multiples of 12 are also multiples of 6 and they all can be written as the sum of nine numbers.

Related questions

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.

The sum of the first 10 multiples of 3 is 165.

975

There are 166 multiples in 500 (500/3=166r2)and 33 multiples in100(100/3=33 r1) if you subtract the later from the first.... 166-33=133

The sum of the first 500 positive integers is: 1 + 2 + 3 + ... + 498 + 499 + 500 = 125250

5 + 10+15+20=50 ans

Sum_ap = ½ × number_of_terms × (first_term + last_term) For the first 88 multiples of three: number_of_terms = 88 first_term = 1 × 3 = 3 last_term = 88 × 3 = 264 → Sum = ½ × 88 × (3 + 264) = 44 × 267 = 11748

4 + 8 + 12 = 24

10+20+30 = 60

Sum of first 25 multiples of 44+8+12....100taking 4 common4(1+2+3+4....+25) = 4*325 =1300

No. To be divisible by 3, if the sum of the digits is divisible by 3 so is the original number, otherwise the remainder when the sum is divided by 3 is the remainder when the original number is divided by 3. 5 + 0 + 0 = 5 5 ÷ 3 = 1 r 2, so 500 ÷ 3 has remainder 2.

3+6+9...+45taking 3 common3(1+2+...+15) = 360