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.
10 is divisible by both 5 and 2, as are any multiples of 10.
The common multiples of 2 and 5 are 10, 20, 30, 40, etc.In detail:A common multiple is a number divisible by both (or all) numbers: 2 and 5, here.Knowing that all numbers divisible by 2 have a 0, 2, 4, 6, or 8 in the ones place combined with the knowledge that all numbers divisible by 5 end with 0 or 5 (excluding 0 in both cases), we can say the only multiples of each number that are common to both are numbers that end with 0.Another way of analyzing this problem is to say that both 5 and 2 are factors. 10 is clearly a multiple of 5 and 2 because 2(5)=10. Any number divisible by 10 is thus also divisible by 2 and 5. All numbers divisible by 10 end in 0. Thus, a number ends in 0 if and only if it is divisible by 2 and 5.Answer: Common multiples of 2 and 5 end with zero.
5 and 25
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.
10, 20, 30, 40, and so on. In other words, any multiple of ten (and only multiples of ten) are divisible by both 2 and 5.
Both 2 and 5 are prime numbers so their product 2 x 5 = 10 is the Lowest (or Least) Common Multiple. The common multiples of 2 and 5 are therefore any number which is a multiple of 10.
Multiples of 50 are the only numbers that are both. All other multiples of 5 aren't.
35 because 5x7 is 35 so that is a multiple of BOTH and 45
There are infinetly many numbers which have 2 and 5 as their factors. Some examples are 10, 20, 30, 40, etc. 2 and 5 both are prime numbers and their common multiples are the multiples of 2 x 5 = 10. So all the multiples of 10 have 2 and 5 as their factors. However if consider 2 and 5 as the only two proper factors, then the required number is 10.
Since both 3 and 5 are prime numbers, only numbers that are multiples of its product are the numbers that are divisible by both. 15 is the LCM of 3 and 5 and hence all multiples of 15 are divisible by both 3 and 5
2, 3 and 5 go into multiples of 30.