Oh, dude, multiples of 5 are like the easiest thing ever. It's like counting by fives, you know? So, you've got 5, 10, 15, 20, and so on, all the way up to 1000. I mean, it's not rocket science... unless you're counting multiples of 5 on a rocket ship, I guess.
5 of them.
Since you didn't specify a single number, and all numbers are multiples of themselves, the five smallest multiples are the counting numbers 1 to 5.
The LCM of two consecutive numbers is their product. The LCM of two consecutive multiples of 5 is their product divided by 5. Two consecutive numbers cannot be multiples of 5.
All numbers do.
99 of them.
There are 199 multiples of 5 in that range. There are 128 multiples of 7 in that range. There are 28 numbers on both lists. 1000 - 299 = 701
5 of them.
the numbers go up in 5, making the Nth Term 5n.
There are 2000 4-digit numbers that are multiples of 5, so, instead of listing them all, it is equally valid to say: Any4-digit number whose final digit is either a 5 or a 0 is a multiple of 5. Get Right? :P
Multiples of 50 are the only numbers that are both. All other multiples of 5 aren't.
Multiples of 5.
All multiples of 5 are numbers ending in either 5 or 0.
The first multiple of 3 is 3; the last multiple of 3 below 1000 is 999. (This can be determined by dividing 1000 by 3, ignoring the remainder, and then multiplying that number by 3 to determine the largest multiple of 3 less than 1000.) Since 999 is 3 times 333, there are 333 multiples of 3 that are less than 1000. So, consider them by pairs: 1st and 333rd = 3 + 999 = 1002 2nd and 332nd = (2 x 3) + (332 x 3) = 6 + 996 = 1002 3rd and 331st = (3 x 3) + (331 x 3) = 9 + 993 = 1002 up to 166th and 168th = (166 x 3) + (168 x 3) = 498 + 504 = 1002 167th = 501 (which is half of 1002) In other words, since 333 is not an even number, there are (333 - 1)/2 = 166 pairs, plus that extra half of a pair. The sum of all the multiples of 3 less than 1000 is 166.5 x 1002 = 166,833. The same can be done for the multiples of 5. The first multiple of 5 is 5; the last multiple of 5 below 1000 is 995. (This can be determined by dividing 1000 by 5, subtracting 1 since it divided evenly and you need the largest multiple less than 1000, and then multiplying that number by 5 to determine the largest multiple of 5 less than 1000.) Since 995 is 5 times 199, there are 199 multiples of 5 that are less than 1000. So, consider these by pairs as well.e are 1st and 199 = 5 + 995 = 1000 2nd and 198 = 10 + 990 = 1000 up to 100th = 500 (which is half of 1000) In other words, since 199 is not an even number, there are (199 - 1)/2 = 99 pairs, plus that extra half of a pair. So, the sum of all the multiples of 5 less than 1000 is 99.5 x 1000 = 99,500. If the desired answer is the sum of all numbers less than 1000 that are either multiples of 3 or 5, then the numbers that are multiples of both 3 and 5 have been included twice - once as multiples of 3 and again as multiples of 5. So, since all numbers that are both multiples of 3 and multiples of 5 are multiples of 15, determine the sum of all the multiples of 15 and subtract it from the sum of the multiples of 3 and the multiples of 5. We can repeat the same procedure again. The first multiple of 15 is 15; the last multiple of 15 less than 1000 is 990. Since 990 is 15 x 66, there are 66 multiples of 15 less than 1000. 1st and 66th = 15 + 990 = 1005 2nd and 65th = 30 + 975 = 1005 and so on Since 66 is an even number, there are 66/2 = 33 pairs. So the sum of all the multiples of 15 less than 1000 is 33 x 1005 = 33,165. The sum of all the multiples of 3 and all the multiples of 5, but not counting them twice, is 166,833 + 99,500 - 33,165 = 233,168.
The common multiples of 6 and 5 are 30, 60, 90, 120, 150, 180, and so on.
the common multiples of 5 and 6 is 30
To find the numbers between 10 and 50 that are multiples of both 3 and 5, we need to find the numbers that are multiples of the least common multiple of 3 and 5, which is 15. The multiples of 15 between 10 and 50 are 15, 30, and 45. Therefore, there are 3 numbers between 10 and 50 that are multiples of both 3 and 5.
There are eleven such numbers.Get the common multiple of those numbers. All other common multiples will be multiples of this common multiple; that is, you can multiply the common multiple by 1, by 2, by 3, etc. to get additional numbers that are multiples of both numbers.