If $81 is to be divided among "n" people, where "n" > 1, so that each gets $x, where x is a whole number > 1, how many different values could there be for n?
2006-09-19 10:05:42 · 7 answers · asked by Allison B 2 in Science & Mathematics ➔ Mathematics
List all the factors of 81
1, 3, 9, 27, 81
Answer: There are 3 possible values for n.
You eliminate 1 because n must be greater than 1, and you eliminate 81 because x must be greater than 1.
3 people would get $27 each.
9 people would get $9 each.
27 people would get $3 each.
2006-09-19 10:10:17 · answer #1 · answered by MsMath 7 · 1⤊ 0⤋
Since n and x must both represent whole numbers - you are looking for all possibilities of n*x = 81, where n and x are whole numbers (and n and x are both larger than 1).
what whole numbers divide evenly into 81 (other than 1?)
Looks to me like 1, 3, 9, 27, and 81.
you can give 3 people $27, 9 people $9, or 27 people $3.
Looks to be me like there are 3 possible values for n.
2006-09-19 17:13:19 · answer #2 · answered by captain2man 3 · 0⤊ 0⤋
You have to divide 81 evenly between n people basically.
And x must be a whole number...
so, n = 3, x = 27
n = 9, x = 9
n = 27, x = 3
3 different values of n.
2006-09-19 17:13:28 · answer #3 · answered by AresIV 4 · 0⤊ 0⤋
81 = 3 x 3 x 3
So the combinations are:
1 x $81
3 x $27 <---
9 x $9 <---
27 x $3 <---
81 x $1
But you said n > 1 and x > 1, so there are only 3 choices meeting your criteria:
n = 3 (x = $27)
n = 9 (x = $9)
n = 27 (x = $3)
for a total of three (3) choices.
2006-09-19 17:10:33 · answer #4 · answered by Puzzling 7 · 1⤊ 0⤋
3 values: 3, 9 and 27.
2006-09-19 17:11:11 · answer #5 · answered by Pascal 7 · 1⤊ 0⤋
Find the factors for 81: 3, 9. Thats it.
2006-09-19 17:09:48 · answer #6 · answered by Jose 2 · 0⤊ 2⤋
it can be 3*27;9*9;27*3
n can have 3 values 3,9,27
2006-09-19 17:12:29 · answer #7 · answered by raj 7 · 0⤊ 0⤋