2006-07-14 02:08:29 · 4 answers · asked by Anonymous in Education & Reference ➔ Primary & Secondary Education
If you want 7 different factors, then it would be the product of the first 7 prime numbers including 1.
1*2*3*5*7*11*13 = 6006
If you don't want to include 1, then it would be
2*3*5*7*11*13*17 = 102102
If you want 7 repeating factors, then it could be 1.
1*1*1*1*1*1*1 = 1
A zero can have infinite number of factors as long as one of the factor is 0.
2006-07-14 08:26:10 · answer #1 · answered by MickMan 2 · 1⤊ 2⤋
24 has 8 factors... 1, 2, 3, 4, 6, 8, 12, 24.
If you don't count the number itself, that's seven factors.
In order to have an odd number of factors, a number must be square. I've checked up to 15^2 and all have either more or fewer than 7 factors, but there may be a bigger number that has 7 factors... I don't know.
2006-07-14 02:15:23 · answer #2 · answered by mathsmart 4 · 0⤊ 0⤋
64
1, 2, 4, 8 (square factor), 16, 32, 64
2006-07-14 04:58:55 · answer #3 · answered by jimbob 6 · 0⤊ 0⤋
128
2^7 or (2*2*2*2*2*2*2)
2006-07-14 03:24:25 · answer #4 · answered by jedi 2 · 0⤊ 0⤋