Multiples of both 2 and 5 are numbers that are divisible by both 2 and 5 without leaving a remainder. To find these multiples, you can simply find the common multiples of 2 and 5. The least common multiple of 2 and 5 is the smallest number that is divisible by both 2 and 5, which is 10. Therefore, multiples of both 2 and 5 include 10, 20, 30, 40, and so on.
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The common multiples of 2 and 5 are numbers that can be divided evenly by both 2 and 5. The common multiples of 2 and 5 are multiples of their least common multiple (LCM), which is 10. Therefore, the common multiples of 2 and 5 are all multiples of 10. Similarly, the common multiples of 2 and 6 are multiples of their LCM, which is 6. Therefore, the common multiples of 2 and 6 are all multiples of 6.
The common multiple for 2 and 5 is 1 cause both of the numbers are prime.
Multiples of 12: 1, 2, 3, 4, 6, 12 Multiples of 5: 1, 5
Since they are both prime, just multiple them together (65) and find multiples of that. Common multiples of 5 and 13 are: 65*1 = 65 65*2 = 130 65*3 = 195 etc
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.