2006-11-30 08:01:48 · 8 answers · asked by TJH 2 in Science & Mathematics ➔ Mathematics
9 = 3*3, and 21 = 3*7, so common multiples must have at least two factors of 3, and one factor of 7; ie they must be a multiple of 3*3*7 = 63. Theres only one - 63.
Or you could just list multiples of 21: 21,42,63,84. Only one is a multiple of 9.
2006-11-30 08:04:06 · answer #1 · answered by stephen m 4 · 1⤊ 0⤋
Multiples Of 21
2016-10-18 11:00:29 · answer #2 · answered by ? 4 · 0⤊ 0⤋
9 = 3*3, 21=3*7. The common multiple less than 100 is 3*3*7=63.
You find this by taking the highest power of each prime number from each number. For the prime factor 3, the highest power is 2, i.e. 3*3; for the prime factor 7, the highest power is 1, i.e. 7. Then multiply the factors together, (3*3)*(7) = 3*3*7 = 63.
2006-11-30 08:04:43 · answer #3 · answered by maegical 4 · 0⤊ 0⤋
63
2006-11-30 08:04:39 · answer #4 · answered by jokidd 1 · 0⤊ 0⤋
63 is the only one I can find under 100.
2006-11-30 08:11:22 · answer #5 · answered by Anonymous · 0⤊ 0⤋
0nly 63
2006-11-30 08:08:04 · answer #6 · answered by raj 7 · 0⤊ 0⤋
I think you should take a break from the math and focus on your spelling and grammer homework. Also you sound very rude for someone who is asking for help.
2016-05-23 05:44:57 · answer #7 · answered by Anonymous · 0⤊ 0⤋
Is that a question? It needs another verb.
2006-11-30 08:07:03 · answer #8 · answered by Anonymous · 0⤊ 1⤋