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by Frankenstein » Wed Jun 22, 2011 10:22 am
Hi,
3(12!)-15(11!) = 3*12*11! - 15*11! = (36-15)*11! = 21*11!
[22(11!)+22!] = [(21+1)(11!)+22!] = 21*11! + 11! + 22!
22! = 11!*12*13*..*21*22 = 11!*21*(12*13...20*22)
So, 22! is divisible by 21*11!
21*11! is divisible by 22*11!
We are left with 11!

So, [(21+1)(11!)+22!] when divided by 21*11! leaves 1*11! = 11!

Hence, A
Last edited by Frankenstein on Sun Jun 26, 2011 9:35 am, edited 2 times in total.
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by joinn » Sat Jun 25, 2011 9:00 am
Hi Frankestein, could you please elaborate, I didn't understand from third step
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by Frankenstein » Sun Jun 26, 2011 9:37 am
joinn wrote:Hi Frankestein, could you please elaborate, I didn't understand from third step
Hi,
There were a couple of typos(wrote 22 instead of 21) in my previous post. I have edited my post now. If you still think it is not clear, let me know where you have problem. I will try to make it better.
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by joinn » Mon Jun 27, 2011 6:10 pm
yeah my doubt is from third step
when 11!*21! divides 22! then what about 22*20*19*18*17*16*15*14*13*12 aren't these also not remainders ?
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by Frankenstein » Mon Jun 27, 2011 8:46 pm
joinn wrote:yeah my doubt is from third step
when 11!*21! divides 22! then what about 22*20*19*18*17*16*15*14*13*12 aren't these also not remainders ?
Hi,
That is the quotient, not remainder.
Example: 16 when divided by 2 gives quotient of 8, but the remainder is zero.
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by winniethepooh » Mon Jun 27, 2011 10:16 pm
Wrong question accompanied with wrong answer!
Try calculating with a calculator!
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