[ankursanghi]

Originally Posted by clsuresh

Hi, Thakarsagar

I think that's not the right answer but I think u got into the right way of thinking. Try on the same lines. All the best and I appreciate u for answering my previous question asking the highest power of 3 in 1! x 2! x 3! x 4! x .......... x 98! x 99! x 100!

Hi Suresh,

I would like to see the explanation on this question (highest power of 3 in .... ). Where is it??

Your efforts to provide us fundoo questions is appreciated!!

[Stefan R]

Hi Suresh,

This is solvable, but I want to know your approach :-)

What is the remainder when 17^19 + 13^19 is divided by 25.

Regards,

[thakarsagar]

Originally Posted by Stefan R

Hi Suresh,

This is solvable, but I want to know your approach :-)

What is the remainder when 17^19 + 13^19 is divided by 25.

Regards,

Ans: 1

(17^19 + 13^19) % 25
= (3 + 23) % 25
= 1

[clsuresh]

Hi Ankur,

Here's the solution

the above series can be written as ,
1^100 * 2^99 * 3^98 * 4^97 *.... * 100^1
Note that for each no n the no of powers will be 101 - n
First we find the multiples of 3...
ie, 3^98, 6^95, 9^92, 12^89,............ , 99^2
So the no. of 3's in the above numbers is
98+95+92+................+2 = 1650...--------------------->(1)
Now, we need to add multiples of 3 contained in multiples of 9....
ie, 9^92, 18^ 83 , 27^74,........... ... ,99^ 2 i.e 92+83+74+........+2= 517 (see, the power of 9 will be 9^92 = (3*3)^92... similarly for 18,27,...,99)-----------------(2)
Now, add multiples of 3 in 27...
ie, 74 + 47 + 20 = 141 --------------->(3)
finally, for 81 we have 20--------------->(4)
Add (1) (2) (3) and (4) we have, 2328...

[Stefan R]

Originally Posted by thakarsagar

Ans: 1

(17^19 + 13^19) % 25
= (3 + 23) % 25
= 1

can't be 1, has to be a multiple of 5, you can see that :-)

infact, it's 3 +2.

Ok, i make it 17^18 + 13^18 by 25. Approach please???

[thakarsagar]

Originally Posted by Stefan R

can't be 1, has to be a multiple of 5, you can see that :-)

infact, it's 3 +2.

Ok, i make it 17^18 + 13^18 by 25. Approach please???

yes, i made a silly calculation mistake..

17^18 + 13^18 % 25
= 9 + 4
= 13...

[Stefan R]

Originally Posted by thakarsagar

yes, i made a silly calculation mistake..

17^18 + 13^18 % 25
= 9 + 4
= 13...

Right answer :-)

[clsuresh]

Hi. Here is our next question.

Find the remainder when 104^303 is divided by 101.

[thakarsagar]

Originally Posted by clsuresh

Hi. Here is our next question.

Find the remainder when 104^303 is divided by 101.

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

[clsuresh]

Thakar,

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