Extremely dissatisfied with the CAT and the way it is being organised . Looking forward to get a good score in GMAT. Require an active participation which could help us all in a long run .

**Today's funda...it's a combination of few of maximus's old posts...****factorial based questions asking no. of zeroes and max power of sum integer****.**

Find the no. of zeroes at the right end of 300!for every zero, we require 10..n every 10 is made up of 5x2.in the expression 1x2x3...300, multiples of 2 wud obviously be more than the multiples of 5...so v need to find the maximum power of 5 in 300!300/5 = 60 (because every fifth no. is a multiple of 5)300/25 = 12(because every mutiple of 25 has two 5s in it) or, 60/5=12300/125 = 3 (because multiples of 125 have three 5s in it) or,12/5 = 2now 2 cannot be further divided by 5 so add all the quotients...60 + 12 + 2 = 74.we might also get the same type of questions in a different form,

**500! is divisible by 1000^n...what is the max. integral value of n? **now every 1000 is made up of 3 5s and 3 2s....2s are redundant...we need to count no. of 5s....so find total no. of 5s and divide by 3500/5 = 100100/5 = 2020/5 = 4100 + 20 + 4 =124124/3 = 41.33max integral value is 41.

**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 answer500/11 = 4545/11 = 4ans 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 = 5050/2 = 2525/2 =1212/2 = 66 /2 = 33/2 = 1add 'em all...97. 97/5 = 19.so the answer wud be 19had v 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...when in dilemma as to which prime no. wud be the deciding factor (e.g. a divisor like 144...its not possible to decide whether 3 or 2 will give the right answer) ....take out answer using both the prime nos...the one thats less is the right answer.

**50! is divisible by 144^n...what is the max. integral value of n?**144 = 2^4 * 3^2...difficult to decide whether 3 or 2 will be the deciding factor...count 2s50/2=2525/2=1212/2=66/2=33/2=1sum=47answer = 47/4 = 11.count 3s50/3=1717/3=55/3=1sum = 2323/2 = 11a tie...else the smaller value wud have been the answer.

**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.

**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 3^2to 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! = 255hence, no. of zeroes = 256/6 = 42.just to check...3s = 126, 126/2 = 63>42ans-42

**Questions based on this concept**400! is divisible by x^n. what is the max. possible integral value of n if the value of x is:

Q1. 300

Q2. 99

Q3. 500

Q4. 320

Q5. 770

Q6. 5200

Q7. 270

Q8. 686

Q9. 338

Q10. 13000