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the expression is 1! *2!*3!*........49!*50!

where * is multiplication and ! is the factorial function

An Unconventional question... which can be solved by multiple logic... what is your take ?Lol.. I do not know.. LOl.. You are blowing my mind... You havent read any of my hubs lately.. I am depressed.. Miss you son

Each factorial above 4 contains the expression containing 5*4, which produces a zero. That's 46 of them.

Each one above 9 has a *10 in it - that's 41 more.

Each one above 19 has a *20 in it - that's 31 more.

Each one above 29 has a *30 in it for 21 more

Each one above 39 has a *40 in it for 11 more

Add one for *50 and you have a grand total of 151 zeros.

I think.Interesting way to calculate. . . . But what about 14x15?, 24x25?, 34x35?, 44x45?

I haven't even started trying to figure this out, so these may already be included in some way I haven't noticed yet. Not sure I'll have time to figure it anytime soon, but I'll be watching this. Fun!

Adding in my head I got 262 trailing zeroes by counting the repetitions of factors of 5.

Since 2 is a smaller prime factor than 5, there will be more 2's than 5's in the prime factorization of the expression. This means there will be more than enough 2's to pair up with all the 5's so there's no need to count them.

What's the correct answer?Prime factors of:

50=2*5*5

54=2*3*3*3

18=2*3*3

49=7*7

point being that in all of these there are fewer 2s than larger factors. Better count the 2s as well.

Still, I like your thinking. Did you forget either the first one (five itself) or the second one in 50? Look above for my most recent guess of 263.Yes, some individual numbers' prime factorizations contain fewer 2s than other primes. But in the aggregate product 1!*2!*...*50! the 2s outnumber the 5s, so it suffices to count just 5s

Think of it like this, 50% of the numbers are divisible by 2, but only 20% are divisible by 5.

It is 262 my friend and your concept is exact.... we need to look the number of 2*5s and clearly... 2s will be more so we just need to look for the number of 5s

I hoped I would find time to really go through this and figure it out, but I came up with 236 (not quite the same method as either of you). That's a preliminary guess, and I'm going to check myself when I can, but it may be Monday before that will be possible!

EDIT: Now I see the ones I failed to count and I agree that 262 should be correct.Rahul I'm mathematically not there. This is not something I can figure out but when exploring mathematical sources this is an explanation I found about zeros.

1.a) 10; b) 50; c) 100 (10^n = one followed by n 0s)

2.a) 9^6 = 59049*9 ~ 540000 < 10^6

b) 3^14 = 59049*81 ~ 4800000 < 10^7

3.a) 7^10 > 6^10 (7>6)

b) 8^10 (note 8^10 = (2^10)^3 = 1024^3 > 10^9 > 10^8)

c) you can't say greater with only one number.

And I couldn't even imagine how to formulate yours but interesting thought. I'll agree with all those who are smarter than me.Whoa...!

I wish I could even understand the points you listed! Lol!

When rahul posted this, I couldn't figure out how to start finding the number of zeroes, but wilderness's post got me started in the right direction. C-G's turned out to be even clearer.

If you think about it, a zero (after the number 1) has to represent the product of a 5 and some even number (therefore a 2 as one of its factors). And so, if you know how many 5's are factors of the numbers in the factorial expressions, you know how many zeroes there will be. No matter how many other prime numbers are factors, there*must*be a 5 (and a 2) for every zero in the final product. And so, if you count the 5's that are factors, as C-G said, you will know how many zeroes you will eventually have. (Every other number in counting is even and thus has a 2 as one of its factors. This means that automatically there are far more 2's as factors than there are 5's.)

There are no 5's as factors in 1! , 2! , 3! , and 4!; [total=0]

there is one 5 as a factor in**each**of these: 5! , 6! , 7! , 8! , and 9!; [total=5]

there are two 5's that are factors in 10! , 11! , 12! , 13! , and 14!; [total=10]

three 5's as factors in 15! , 16! , 17! , 18! , and 19!; [total=15]

four 5's are factors in 20! , 21! , 22! , 23! , and 24!; [total=20]

Total so far = 50 5's (and therefore 50 zeroes to this point)

Then, because 25 = 5x5, all of the factorial expressions from 25! through 49! will have one additional 5 as a factor, and 50! will have two additional 5's as factors.

There are six 5's as factors in 25! , 26! , 27! , 28! , and 29!; [total=30]

seven in 30! , 31! , 32! , 33! , 34!; [total=35]

eight in 35! , 36! , 37! , 38! , 39!; [total=40]

nine in 40! , 41! , 42! , 43! , 44!; [total=45]

and ten in 45! , 46! , 47! , 48! , 49!; [total=50]

and twelve in 50!. . . . . . .

...for a grand total of 262 zeroes.Now you have got it bang on Aficionada.... U explained it really well

Now how about knowing the fact that this question was one of many saked in the prestigious CAT exams of India... and you get a maximum of a minute to solve it

Can u imagineindeed!

I really canimagine!**not**

More power - and good luck - to everyone who is taking the CAT and similar exams around the world!

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