SOLUTION: i am the number which will make a rectangle 3 times wide, i am the number which has exactly 8 factors, i am the number which has one of my factor as 4, i am the number which is not

Algebra ->  Rectangles -> SOLUTION: i am the number which will make a rectangle 3 times wide, i am the number which has exactly 8 factors, i am the number which has one of my factor as 4, i am the number which is not      Log On


   



Question 983976: i am the number which will make a rectangle 3 times wide, i am the number which has exactly 8 factors, i am the number which has one of my factor as 4, i am the number which is not a multiple of 5 or 7.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
"I am the number which will make a rectangle 3 times wide" may mean that
if the number is the length of a rectangle,
it will be twice the whole number measurement of the width of that rectangle.
In other words, I take that to mean that the number is a multiple of 3 .
That means 3 is a factor of the mystery number.

"I am the number which has one of my factor as 4" means that 4=2%5E2 is a factor of the mystery number.

"I am the number which is not a multiple of 5 or 7" tells us that 5 and 7 are not prime factors of that mystery number, but it is not a particularly useful clue.
Other prime numbers could be factors, like 11, 13, 17, 19, 23, etc .

"I am the number which has exactly 8 factors" is an important clue.
A number N that has 4=2%5E2 and 3 , but not 5 or 7 as factors
has a prime factorization of the form
N=2%5Ea%2A3%5Eb%2A11%5Ec%2A13%5Ed%2A17%5Ee%2A%22...%22 , with a%3E=2 and b%3E=1.
The factors of N , with their prime factorizations, range from
1=2%5E0%2A3%5E0%2A11%5E0%2A13%5E0%2A17%5E0%2A%22...%22 to N=2%5Ea%2A3%5Eb%2A11%5Ec%2A13%5Ed%2A17%5Ee%2A%22...%22 .
There are
a%2B1%3E=2%2B1=3 different possible exponents for 2 ,
b%2B1.=1%2B1=2 different possible exponents for 3 ,
c%2B1 different possible exponents for 11 ,
d%2B1 different possible exponents for 13 ,
and so on.
With all the combined possible choices, we can make
%28a%2B1%29%2A%28b%2B1%29%2A%28c%2B1%29%2A%28d%2B1%29%2A%28e%2B1%29%2A%22...%22 factors.
If N has exactly 8=4%2A2 factors,
knowing that a%2B1%3E=3 and b%2B1%3E=2 ,
it is obvious that it must be
a%2B1=4<--->a=4-1=3 ,
b%2B1=2<--->b=2-1=1 ,
c%2B1=1<--->c=0 ,
d%2B1=1<--->d=0 ,
e%2B1=1<--->e=0 , and so on.
So, N=2%5E3%2A3%5E1%2A11%5E0%2A13%5E0%2A17%5E0%2A%22...%22=2%5E3%2A3=8%2A3=highlight%2824%29 .