[thakarsagar]

Originally Posted by clsuresh

Thakar,

Well done, Can u please tell me the way in which u answered. ( did u use FERMAT's rule)

yes...suresh, my name is sagar...



Am tired of remainder problems...here is a different one...

i have infinite no of 1 Re, 5 Re, 10 Re and 25 Re notes...in how many ways can i give the change for Rs 100???
-sagar

[Oxymoron]

Could junta who answer here also elaborate on how they arrived at the answer..that would make following the train of thoughts that much easier...I know it's just a teeny weeny extra effort in typing out a few lines extra but could make the process of understanding that much easier and smoother..

Girish...!

[clsuresh]

Sagar,

Is it 78, Please confirm

[thakarsagar]

Originally Posted by clsuresh

Sagar,

Is it 75. Please confirm

No suresh, i know the answer for this....but i have not been able to derive it fully...i thought on the following lines...

We need to find it for all the subsets of 1 5 10 and 25 ie 16-1 sets...

using anyone of 1 , 5 10 and 25 -> 4 ways
using 1 and 5 -> 19C1 ways...(its equivalent to solving the equation a + 5b = 100...Now this is limited by the highest coefficient...so divide 100/5 = 20...now any of this 20 i can form a partition...one part been paid by 5 rupee notes and other by 1 rupee notes...)
using 1 and 10 -> 9C1 ways...
using 1 and 25 -> 3C1 ways ...
.
.
.
See, if you can work this ways...
-sagar

[clsuresh]

sagar,

I think that is going to be hectic. The way in which i solved is forming only a subset of the answer. Just give me sometime I will come out with a faster approach.

[federerfan]

hi suresh and sagar ..... thanks a lot for giving us such wonderful ques. but it wud be of a gr8 help if u can give ur approach in solving them .

[clsuresh]

sagar,

Is it 229. confirm.........

[Ashish_4_cat]

Originally Posted by thakarsagar

Ans: 27...welcome back suresh...
-sagar

sagar can u elobrate on how u got the answer
i am gettin (3)^303
after that i am getting stuck up
wat 2 do next????

[clsuresh]

Dear Students,

I have posted a question earlier asking the remainder of 104^ 303 when divided by 101 and our friend Sagar answered it using the FERMAT's RULE.

Here is the rule.

Let P be any prime number and N be a number which is not divisible by P.

Then the remainder obtained when N^(P-1) divided by P is 1.
So applying this rule here ( since 101 is prime) and 104 is not divisible by 101,
we can write the given problem as

(104^100)^3 x (104^3)

and (104^100)^3 leave a remainder 1 according to Fermat's rule. (i.e 104^100 gives rem 1)

So now the problem finally becomes 104^3 which leaves remainder 27 when divided by 101.
I hope I am clear.

[clsuresh]

Hi friends,

Here comes our next question.
Find the total number of ways in which two sqares (1x1) can be selected from a chess board such that they will have a side in common.