Q: The sum of all the multiples of 3 or 5 below 1000?

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That would be equal to the sum of all positive integers less than 1000 minus the sum of all positive multiples of three that are less than 1000. That would be equal to: (1000 + 1) * (1000 / 2) - 3 * (333 + 1) * (333 / 2) = 1001 * 500 - 1000 * 166.5 = 500500 - 166500 = 334000

All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.

All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.

There are infinitely many multiples of 9 and it is not possible to add them all.

The sum is 10,000

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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.

That would be equal to the sum of all positive integers less than 1000 minus the sum of all positive multiples of three that are less than 1000. That would be equal to: (1000 + 1) * (1000 / 2) - 3 * (333 + 1) * (333 / 2) = 1001 * 500 - 1000 * 166.5 = 500500 - 166500 = 334000

69342

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.

All multiples of 3 have digits that add up to a multiple of 3. There are 333 multiples of 3 between 1 and 1000.

There are infinitely many multiples of 9 and it is not possible to add them all.

Well, you could add them one by one, or write a small computer program to do that, but here is a shortcut:Use the formula for an arithmetic series, to find the sum of all the multiples of 3 (3, 6, ... 999).Similarly for all the multiples of 5 (5, 10, ... 995).Add it all up.Multiples of 15 will be counted twice in his calculation, so you'll have to calculate the sum of all the multiples of 15, and subtract it from the above.Well, you could add them one by one, or write a small computer program to do that, but here is a shortcut:Use the formula for an arithmetic series, to find the sum of all the multiples of 3 (3, 6, ... 999).Similarly for all the multiples of 5 (5, 10, ... 995).Add it all up.Multiples of 15 will be counted twice in his calculation, so you'll have to calculate the sum of all the multiples of 15, and subtract it from the above.Well, you could add them one by one, or write a small computer program to do that, but here is a shortcut:Use the formula for an arithmetic series, to find the sum of all the multiples of 3 (3, 6, ... 999).Similarly for all the multiples of 5 (5, 10, ... 995).Add it all up.Multiples of 15 will be counted twice in his calculation, so you'll have to calculate the sum of all the multiples of 15, and subtract it from the above.Well, you could add them one by one, or write a small computer program to do that, but here is a shortcut:Use the formula for an arithmetic series, to find the sum of all the multiples of 3 (3, 6, ... 999).Similarly for all the multiples of 5 (5, 10, ... 995).Add it all up.Multiples of 15 will be counted twice in his calculation, so you'll have to calculate the sum of all the multiples of 15, and subtract it from the above.

They are infinitely many and they form an increasing sequence the sum is infinite.

-3

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

The sum of three consecutive multiples of 6 is 666, the multiples are 216, 222 and 228.