2006-09-22 03:53:23 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
15=3*5*1
21=3* 1*7
LCM=Least common multiplier=3*5*7=105.
So,5 smallest numbers that are multiples of both 15 and 21
=105*1,105*2,105*3,105*4,105*5
=105,210,315,420,525
2006-09-22 04:13:36 · answer #1 · answered by Anonymous · 0⤊ 0⤋
15=3x5 21=3x7 so the smallest number divisible by both 15 and 21 is 3x5x7=105. The five smallest numbers divisible by 21 and 15 are the five smallest multiples of 105 or 105,210 315., 420 and 525.
2006-09-22 11:07:56 · answer #2 · answered by True Blue 6 · 0⤊ 0⤋
For getting this, the smallest number which is the multipleof both 15 and 21 will be the least common multiple or LCM of the two numbers. i.e., 105
the other numbers will be simply the multiples of 105 i.e., 210, 315, 420, 525, 630.
2006-09-22 12:38:30 · answer #3 · answered by jammy 2 · 0⤊ 0⤋
The smallest number multiple of 15 and 21 is their LCM 105
so smallest 5 numbers are 105, 105*2,105*3.105*4,105*5
2006-09-22 11:14:48 · answer #4 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
21=7*3
15=5*3
So minimum number divisible by 21 & 15 = 5*7*3 = 105
The other numbers are multiples of 105
the numbers are {105,210,315,420,525}
2006-09-22 10:57:14 · answer #5 · answered by astrokid 4 · 0⤊ 0⤋
105, 210, 315, 420, 525
2006-09-22 10:58:21 · answer #6 · answered by iyiogrenci 6 · 0⤊ 0⤋
15 = (1,3,5,15)
21 = (1,3,7,21)
so 1) (3*5*7) = 105 ....(2)210....(3)315....(4) 420 .... (5) 525
2006-09-22 11:48:57 · answer #7 · answered by Brian D 5 · 0⤊ 0⤋
15 = (1,3,5,15)
21 = (1,3,7,21)
2006-09-22 23:33:42 · answer #8 · answered by Huda H 2 · 0⤊ 0⤋
105,210,315,420,525
2006-09-22 10:58:27 · answer #9 · answered by hidayah tarmizi 2 · 0⤊ 0⤋