If I multiply all whole numbers from 1 through 100, the largest power of 4 that is a factor of the product is A: 4^25 B: 4^32 C: 4^48 D: 4^50
Where do you get these questions? -.-
Lol. From a contest I took this year.
Sadly, I failed the contest. Contest too hard. T~T
hint: list all the multiples of 4 from 1 to 100
LIST all them multiples of from 1 to 100? There's 25 number that is a multiple of 4.
LOL Okay, annoying...
I'm quite certain of my answer, but explaining it... gah...
Say we have the product from 1 to 100 of all integers...
you'll have 25 multiples of 4 you'll also have 50 multiples of 2....they pair up (in any way you want) to have 25 pairs which are multiples of 2*2 = 4 so you really have 25+25 = 50 copies of 4 being multiplied
Yep.
In fact, let's have it in prime factors...
and now, let's take away all the 2's Actually, we need to count how many times 2 is multiplied in the factorisation of the product of all integers from 1 to 100.
@jim_thompson5910 are you accounting for the fact that half the multiples of 2 are already accounted for as multiples of 4? And that a factor of 2 is only 4^(1/2)? So shouldn't we have: 25 multiples of 4 50 multiples of 2 - 25 multiples of 4 already counted = 25 new multiples of 2 = 12.5 multiples of 4?
oh true
If I absolutely had to guess I would say C. I'm honestly not sure how I would solve this problem.
so (multiples of 2) + (multiples of 4) - (multiples of 2 and 4) = (25)+(25) - (25) = 25
So... there are 50 multiples of 2, so that gives us 50 2's. There are 25 multiples of 4, so that gives us a FURTHER 25 2's There are 12 multiples of 8, so that gives us a FURTHER 12 2's There are 6 multiples of 16, so that gives us a FURTHER 6 2's There are 3 multiples of 32, so that gives us a FURTHER 3 2's There is 1 multiple of 64, so that gives us a FURTHER 1 2. Add those up, you get 50+25+12+6+3+2 = 96 2's So \[\Large 2^{96}=4^{48}\]is your man :D
I think he nailed it
Wowow. You too good TJ.
It happens :D
bow to TJ* I want to be your student. :3
LOL that'll be the day :D
:D
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