Simply calculate 3578 x 1, 3578 x 2, 3578 x 3, etc.
Multiples of 6 must also be multiples of 1, 2 and 3.
A normal die has two multiples of 3: 3 and 6.
Multiples of 3 are 3, 6, 9 and so on. Multiples of 7 are 7, 14, 21 and so on. The common multiples of 3 and 7 include 21, 42, 63 and so on.
2, 3 and 5 go into multiples of 30.
To get the multiples of 6, calculate: 6 x 0 6 x 1 6 x 2 6 x 3 etc. Or add 6 to one of the multiples, to get the next one.
By making a table of multiples and finding when the multiples for every number is the same.
Simply calculate 3578 x 1, 3578 x 2, 3578 x 3, etc.
The multiples are 89,178,267,356,445,534,623 and calculate it by add 89 and click equals and it will keep going.
3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99,102All the multiples of 3.
Multiples of 353 include 353, 706, 1059, 1412, 1765, and so on.
Multiples of 6: 1,2,3,6 Multiples of 4: 1,2,4 Multiples of 3: 1,3
the multiples of 3 are........0,3,6,9,12,15,18,21,24,27.... and more
They are the multiples of 3*8 = 24
3,6,9,12,15,18,21,24,27,30,33,36,39,42are multiples of 3 5,10,15,20,25,30,35,40,45,50,55,60,65,are multiples of 5
No - alternate multiples of 3 are odd, and alternate multiples are even.
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.