2007-09-01 11:04:07 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
128=2^7 (2x2x2x2x2x2x2)
129=3x43
105=3x5x7
Basically try to find prime dividors to the number, starting from 2,3,5,7 and so on. The first one that divides the number without a remainder is 100% a factor of that number. Continue that process until the number cannot be divided by anything anymore - that by the way is also a factor of the original number.
Take the third example:
is 105 divisible by 2? - NO
by 3? - YES! 3x35=105 so 3 is one of the factors.
Continue on:
is 35 divisible by 2? - no
by 3? -no
by 5? - YES - 35/5 = 7
and 7 is prime so it joins the list!
It's tat simple.
2007-09-01 11:18:07 · answer #1 · answered by Anonymous · 0⤊ 0⤋
128 = 1, 128 and 2, 64 and 4, 32 and 8, 16
129 = 1, 129 and 3, 43
105 = 1, 105 and 3, 35 and 5, 21 and 15, 7
2007-09-01 18:14:32 · answer #2 · answered by Angela S 2 · 0⤊ 0⤋
128 = 2^7
129 = 3*43
105 = 3*5*7
2007-09-01 19:33:28 · answer #3 · answered by steiner1745 7 · 0⤊ 0⤋
128: 2^7
129: 3 * 43
105: 3 * 5 * 7
2007-09-01 18:15:34 · answer #4 · answered by llamaboy007 2 · 0⤊ 0⤋
128=2^7
129=3*43
105=3*5*7.
2007-09-01 18:32:37 · answer #5 · answered by Anonymous · 0⤊ 0⤋
128
1 * 128
2 * 64
4 * 32
8 * 16
129
1 * 129
3 * 43
105
1 * 105
3 * 35
5 * 21
7 * 15
2007-09-01 18:23:54 · answer #6 · answered by The Glorious S.O.B. 7 · 0⤊ 0⤋