10*9*8*7 should do it because
the rest have factors included in those
dtermine the smallest positive integer that is divisible by each of the first ten numbers (1,2,3,4,5,6,7,8,9,10).
2 answers
9*8*7*5 is enough. 8*5 has the 5 and the 2 needed for 10.
To do this problem in an organized way try facoring each number
1 = 1
2 = 2
3 = 3
4 = 2*2
5 = 5
6 = 2*3
7 = 7
8 = 2*2*2
9 = 3*3
10 = 5*2
so
I need three twos for 8
I need two threes for 9
I need a five for 10 (already have the 2)
I need a seven for seven
I already have the fctors for one through four.
so
2*2*2 * 3*3 * 5 * 7
= 8 * 9 * 5 * 7 = 2520
To do this problem in an organized way try facoring each number
1 = 1
2 = 2
3 = 3
4 = 2*2
5 = 5
6 = 2*3
7 = 7
8 = 2*2*2
9 = 3*3
10 = 5*2
so
I need three twos for 8
I need two threes for 9
I need a five for 10 (already have the 2)
I need a seven for seven
I already have the fctors for one through four.
so
2*2*2 * 3*3 * 5 * 7
= 8 * 9 * 5 * 7 = 2520
1 answer