This post is a just a saved .doc file which I had, and I am merely reproducing it in its original form, I am not sure of its original author/source but I am grateful to him/her/them for the same.
256! is expanded and expressed in base 576 . how many zeroes will this expression have on its right end?
Such questions are same as finding maximum power of 576 in 256!
576 = 2^6 x 3^2
to get six 2s i have to travel eight places...1x2x3x4x5x6x7x8 has seven 2s. but to two 3s i have to travel only six places...1x2x3...6 has two 3s...hence 2 will be the constrain.
total 2s in 256! = 255
hence, no. of zeroes = 256/6 = 42.
just to check...3s = 126, 126/2 = 63>42
ans-42
300! is divisible by (24!)^n. what is the max. possible integral value of n?
such questions are tricky...when u expand 24!...u get 1x2x3...24.
in this range the highest prime no. is 23...so maximum power of 23 in 300! will decide the max value of x...
when v expand 300!...v get a 23 in 23, 46,69,92....
total no of multiples of 23 in 300! will be 300/23 = 13,
forget the fractional part. so the maximum possible answer is 13. hope am clear...else, feel free to revert.
500! is divisible by 99^n...what is the max. integral value of n?
now every 99 is made of two 3s and one 11. obviously 11 will be the deciding factor. so count no. of 11s for the answer
500/11 = 45
45/11 = 4
ans will be 49.
so in such questions, just check which prime no. will be the deciding factor and count the no. of times it occurs. but please understand that highest prime no. is not necessarily always the deciding factor. see this example:
100! is divisible by 160^n...what is the max. integral value of n?
now 160 = 2^5 * 5^1. now although 5 is the biggest prime no. that 160 is made of, the deciding factor wud be 2. because five 2s occur less often than one 5 does. so we'll count the no. of 2s and divide by 5.
100/2 = 50
50/2 = 25
25/2 =12
12/2 = 6
6 /2 = 3
3/2 = 1
add 'em all...97.
97/5 = 19.
so the answer wud be 19
had we taken 5 as the deciding factor, the answer wud have been 100/5 + 100/25 = 24 which is more than 19...hence a wrong answer...
Though the answer reached for 300!/(24!)^n is correct, the approach has a flaw.
ReplyDelete24! has 22 powers of 2 which has the least frequency than any other prime powers in it. So it will be the limiting factor. In fact 24! has 2^22, 3^10, 5^4, 7^3, 11^2, 13, 17, 19, 23. And here 23 is more* frequent than 3^10 which in turn more than 2^22. Answer will be the same 13.
Try finding highest power of 10! in 21!. Answer will be 2 not 3 as obtained by authors approach.
"such questions are really tricky" ;)
"that highest prime no. is not necessarily always the deciding factor" Correct. This was the case here.
*Frequency of 3^10 is less than 23 just for one no. that is 23! which has 1 twenty-three and 9 threes. For 24! to 26! it is equal and 27! onwards 3^10 is more frequent than 23^1. It holds even at 46!, 69! or 529!.
Sumi
Absolutely ....
ReplyDeleteAs I've mentioned in the start of this post, this is not written by me.
I am really sorry for the incorrect procedure mentioned, and I will try my level best to ensure that even if I am posting something written by some-body else, it must be correct.
Thanx. for pointing out the flaw.
hi rohit! could u pls explain as to how in your very first example of expressing 256! in base 576, the ans 42 is the no. of zeros at the end ,according to me it is simply the maximum power of 576 in 256!
ReplyDeleteHi,
ReplyDeleteKindly re-read the post, you are stating the same as written in the post.
The question was to find the no. of zeroes at the end of 256! when written in the base 576. and it IS simply the maximum power of 576 in 256! as stated by you.
hey!
ReplyDeleteok, let me be more specific,how is the maximum power of 576 in 356! the same as the no. of zeros at the end of 256 !?
Hi,
ReplyDeleteSince 256! is a number written in base 10, so the highest power of any number in it will be the same as highest power of 10 if 256! is written in the base of that number.
Let me show this with the help of 2 examples.
Highest power of 3 in 4! = 1.
Now 4!=24, when written in base 3 = 220, so number of zeroes in 220 = 1.
Highest power of 3 in 6! = 2.
Now 6!=720, when written in base 3 = 222200
number of zeroes in 222200 = 2.
Hope it helps.
hey how can i find the highest power of 100! in 10000!
ReplyDelete