Confused on how to apply THe Unique factorzation Theorem

[mr_coffee]

I'm having troubles applying this theorem. IT states Given any integer n > 1, there exist a postive integer k, distinct prime numbers p1,p2,...,pk, and positve integers e1,e2,...ek such that
n = p1^(e1)p2^(e2)p3^(e3)...pk^(ek), and any other expression of n as a product of prime numbers is identical to this except, perhaps, for the order in which the factors are written. The book gives 1 example,
1176 and the answer in the back is, (2^3)(3)(7)^2. I have no idea how they figured this out. Is there a trick or can someone explain that theorem better?

Thanks!

I am trying to figure out 3675, also I'm not allowed to use a calculator.

Also I'm trying to figure out how many 0's are at the end of (45^8)(88^5), the answer is 8 zeros because i used the calculator. But he said u can do it without, the book says, hint: 10 = (2)(5). 45 = (9)(5) = (3)(3)(5); 88 = (22)(4) = (22)(2)(2) but I'm not seeing how this is helping me much

 

[vsage]

for your first question: there is no trick that I know of beside in some cases looking at the properties of the digits and determining from that what divides it evenly (for example any number whose digits sum to a multiple of 3 are divisible by 3) but I doubt that's the route they take.

The method is simple: Check to see if the number is divisible by 2 and then divide it by 2 if yes, if so then check to see if it's divisible by 2 and if so divide by 2 and repeat this until it is no longer divisible by 2. After this, repeat the same process with the next lowest prime number until you get a result of 1 from the original number. Obviously then all the numbers you divided through by are the factors of the original number

As for the second question, look very closely at how many times you can pair certain factors together in (45^8)(88^5) to produce numbers with 0's on the end when the two halves to the pair are multiplied together.

 

[HallsofIvy]

I'm having troubles applying this theorem. IT states Given any integer n > 1, there exist a postive integer k, distinct prime numbers p1,p2,...,pk, and positve integers e1,e2,...ek such that
n = p1^(e1)p2^(e2)p3^(e3)...pk^(ek), and any other expression of n as a product of prime numbers is identical to this except, perhaps, for the order in which the factors are written. The book gives 1 example,
1176 and the answer in the back is, (2^3)(3)(7)^2. I have no idea how they figured this out. Is there a trick or can someone explain that theorem better?

Thanks!

I am trying to figure out 3675, also I'm not allowed to use a calculator.

Try dividing. You should be able to see easily that 3676 is not even and so not divisible by 2. It's not too hard to see that 3 goes into 3675 1225 times so 3675= 3(1225). It's also not hard to see that 3 does not divide into 1225 (when you try you get 408 with 1 left over). The next larger prime is 5 and 1225= 5(245) so 3675= (3)(5)(245). 245 = 5(49) so 3675= 3(5)(5)(49). Of course, 49= 7² so
3675= (3)(5²)(7²).

Also I'm trying to figure out how many 0's are at the end of (45^8)(88^5), the answer is 8 zeros because i used the calculator. But he said u can do it without, the book says, hint: 10 = (2)(5). 45 = (9)(5) = (3)(3)(5); 88 = (22)(4) = (22)(2)(2) but I'm not seeing how this is helping me much

The point is that 10= 2(5)!

Yes, 45= 3²(5) and 88= 22(2)(2)= 11(2³) (Actually, it's easier to see that 88= 8(11) and 8= 2³).

Then 45⁸= 3¹⁵(5⁸) and 88⁵= 11¹⁰2¹⁵.

Multiplying those together shows that (45⁸)(88⁵)= (3¹⁵)(5⁸)(11¹⁰)(2¹⁵). We can put 8 of those 2's together with the 8 5's and see that we have (10⁸) time other things. It's those 8 10's that give the 8 0's at the end of the number.

 

[mr_coffee]

Ahh thanks guys for the help and explanation Ivey. I understand you up to the part where you say,

"We can put 8 of those 2's together with the 8 5's and see that we have (108) time other things."

When you say the 8 5's, do u mean 5^8, and i see 2^15, if so, how did you know to choose those 2 numbers that would produce the 0's at the end? Its true (5^8)(2^10) = 4E8, which will give you 8 0's which is what i wanted. It seems like common sense to you but I'm not seeing it.