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Mathematics 27 Online
OpenStudy (anonymous):

Given that a, b, and c are positive integers, solve the following equation. a!b! = a! + b! + 2^c

OpenStudy (anonymous):

no guessing!

OpenStudy (anonymous):

the answer gives small numbers, so you can guess it

OpenStudy (anonymous):

but that is not fair :-)

OpenStudy (anonymous):

wait I want to solve it algebraically

OpenStudy (anonymous):

good good, hint: start by dividing with b! or a!

OpenStudy (anonymous):

assume that a>=b (without the loss of generality) now we divide by b! a!=a!/b!+1+2^c/b! As each term of the RHS is an integer RHS>=3 this was the second hint :-)

OpenStudy (anonymous):

Well, the good thing is that the only time 2^c/b! can be an integer is if b! is a power of 2. The only time b! is a power of 2 is when b!=2, and so b=2. Now that b!=2, we can see that a!=a!/2+1+2^c/2. We can see that a!-a!/2=a!/2, so a!/2=1+2^c/2. Take out the 1/2 and we're left with a!=2+2^c.

OpenStudy (anonymous):

nearly but b=2 or b=1

OpenStudy (anonymous):

? I got b=2

OpenStudy (anonymous):

for b=1it does not make sense as a!=a!+1+2^c so b=2

OpenStudy (anonymous):

a!/2 = 1 + 2^(c-1)

OpenStudy (anonymous):

OHHOHO I see

OpenStudy (anonymous):

is a is greater than 3 a!/2 is even that cannot be so

OpenStudy (anonymous):

a!/2-1 has to be even

OpenStudy (anonymous):

so a has to be 3

OpenStudy (anonymous):

right nice

OpenStudy (anonymous):

c=2

OpenStudy (anonymous):

yup a=3 b=2 c=2

OpenStudy (anonymous):

I liked this one too :-)

OpenStudy (anonymous):

This one was challenging but it was possible (with hints :)

OpenStudy (anonymous):

these are 1 level more complex, should I look something from tier 4? (hardest)

OpenStudy (anonymous):

lol go for it

OpenStudy (anonymous):

I hardly understand the questions :)

OpenStudy (anonymous):

The divisors of a positive integer, excluding the number itself, are called the proper divisors . If the sum of proper divisors is equal to the number we call the number perfect. For example, the divisors of 28 are 1, 2, 4, 7, 14, and 28, so the sum of proper divisors is 1 + 2 + 4 + 7 + 14 = 28. The first eight perfect numbers are 6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128. Prove that P is an even perfect number iff it is of the form 2^(n-1) * (2^n -1) where 2^n -1 is prime.

OpenStudy (anonymous):

these are sick :-) I have no clue how to do this

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