CAT4MBA

Ans to Q1.

 Reminder of [ (6!)^7! ]²⁶⁶⁶  when divided by 13

= Reminder of [ (720!)^7! ]²⁶⁶⁶  when divided by 13

= Reminder of [ (5)^7! ]²⁶⁶⁶  when divided by 13

= Reminder of [ (25)^7!/2 ]²⁶⁶⁶  when divided by 13

= Reminder of [ (-1)^7!/2 ]²⁶⁶⁶  when divided by 13

As 7!/2 and 2666 is even number the reminder is 1

Notes: for an odd number in the power =>reminder -1 =>reminder 12

.what will b d remainder when [(6!)^7!]^26666 is divided by 13?

1.0  2.12  3.10  .4.1  5.3

ans z 4 i.e remainder will be 1

as already discussed various techniques in my post related to solving remainder questionss

we try to find out remainder 1 or -1

6! can be written as 12*12*5

so when 12 will be divided by 13 remainder will be -1

so only term left is 5, so when 5 will be divided by 13 , remaindr will be 5

so question becomes (5^7!)^26666 .................(1)

so now again using that funda i.e making remainder 1 oe -1

so 7! can be written as 7*6*5*4*3*2*1

so equation 1 becumes (25^7*6*5*4*3*1)^26666     ..............{ 2has been removed to make power of 5  }

so when 25 will be divided by 13 remainder wil be -1

(-1^7*6*5*4*3*1)^26666 remainder is 1

feel free to revert

2nd question will post soon

2.if N b the number of consecutive zeros at the end of the decimal representation of the expression 1! *2!*3!.............................................*99! *100!. find the remainder when N is divided by 1000.

Answer:

First part of the question (and the toughest part):  How many consecutive zeros are there at the end of the decimal representation of the expression 1! *2!*3!.............................................*99! *100!

We know for N! the number of zeros = [N / 5] + [N / 5²] + [N / 5³] + .  .  .  .  because number of 2's always exceed it.

For this question there wont be any term in above equation for 5³ . As the maximum value is 100 here and [100/125] = 0.

So we have to find number of zeros for 5 and 5^2.

For 5

100! 'll give 100/4 = 20 zeros

For numbers 95-99 we 'll get 19 zeros in each case 

For numbers 90-94 we 'll get 18 zeros in each case

 . . . .  . .

Total number of zeros = [ 1 + 2 + 3. . . . . 18 + 19] x 5 + 20

= 970

For 5²

100   ->4

75 - 99 -> 3 x 25

50 - 74 -> 2 x 25

25 - 49 -> 1 x 25

Total number of zero's = 154

Total number of zero's for 5 and 25 =  970 + 154 = 1124

Second part of the question : what is the reminder when 1124 is divided by 1000

ANS: 124

hi Anita...gr* explanation.....10 out of 10....i wz also doin da same...!!!! but wz bit lazy 2 write all da stuff...gr8 approach

 beware Anita...

We know for N! the number of zeros = [N / 5] + [N / 5²] + [N / 5³] + .  .  .  .  because number of 2's always exceed it

in this case this explanation is write...in some questions there are fundas in which 5 or some other prime nos exceed the 2 or some other priime noos...so in that case you have to calculate both the prime nos...

i'l post an example of that question soon

Good point

I should have mentioned it properly and instead of number of 2's always exceed it , i shd have written most of the times number of 2's  exceed  5's

Question 1. What will be the remainder when [(6!)^7!]^26666 is divided by 13?

1. 0 
2. 12 
3. 10 
4. 1 
5. 3

Solution:

(7!) = 5040 = 12 x 420

(7!) x (26666) = 12 x (420 x 26666)

So, [(6!)^(7!)]^26666 = [(6!)]^{(7!) x 26666} = [(6!)^12]^(420 x 26666)

Now, we know from Euler's Theorem that:

If 'p' be a prime number and 'a' be a natural number not divisible by 'p', then (a)^(p - 1), when divided by 'p', the remainder is 1.

Now, note that in the given problem, p = 13 and a = 6! = 720, which is not divisible by p = 13 and hence, by applying Euler's Theorem, we get:

[6!)^12] when divided by 13 leaves a remainder 1.

Hence, [(6!)^12]^(420 x 26666), when divided by 13, will leave a remainder (1)^(420 x 26666) = 1.

So, when [(6!)^7!]^26666 is divided by 13, the remainder is 1.

Thank You.

Ravi Raja

THANX A LOT MY DEAR FRIEND FRIEND.

THANX ravi 4 ur help.

thax a lot dear Anita.i really appreciate ur approch. could u plz go through another 2 questions i hv sent on set theory? pl z go through them & post me d solutions as soon as possible.