[chetna]

Originally Posted by cprash_aggarwal

The answer to this problem is 10
But i m not able to get two extra 2's besides 2^8.

arre simple
we get
8 powers from 10!
n 2 power from 4
8+2=10....our answer

[witchdoctor]

wrong thread

[vikram_k51]

These r the questions frm AimCat0718:

1.Remainder when 90^91 is divided by 13?

a.0 b.7 c.12 d.1

2.Last digit of the LCM of(3^2003-1) and (3^2003+1) :

a.8 b.2 c.4 d.6


For 1st one the 91=13*7=n*(13)

So by Fermat's Little theorem is the remainder not 1?Since 2 and 13 are co-primes.:neutral:

[koolbuddy]

@Cprash... Indeed the answer is 10.
Here is the explanation...
Dekho...
given term = 10!( 1 + 11 + 11.12 + 11.12.13 .....) = 10! * 12 (1 + 11 + 11*13 + ...)
= 10! * 12 (.....)
Now power of 2 in 10! is 8 as solved by others and... power of 2 in 12 is 2.
So total power of 2 in above term is 8 + 2 = 10.
Hamein lagta hai ki hamne sahi jawab diya...
Agar koi truti rah gayi ho to... pls raise the doubt...

Cheers


PS : Edited...

[kingnitin]

Most of you may have seen this question ...and I am posting this with the solution


Is 2222^5555 + 5555^2222 divisible by 7?

Now I want to understand how is 2222^5555 + 5555^2222 = 3^5555 + 4^2222 (mod 7) ????
and 2222^5555 = 3^5 (mod 7) = 5(mod 7).????

Can neone throw more light on this????

The question and its solution using fermet's theorem is available at :

http://www.cut-the-knot.org/blue/examples.shtml

Thanks
NItin

[ashish banaya..]

Originally Posted by kingnitin

Most of you may have seen this question ...and I am posting this with the solution


Is 2222^5555 + 5555^2222 divisible by 7?

Now I want to understand how is 2222^5555 + 5555^2222 = 3^5555 + 4^2222 (mod 7) ????
and 2222^5555 = 3^5 (mod 7) = 5(mod 7).????

Can neone throw more light on this????

The question and its solution using fermet's theorem is available at :

http://www.cut-the-knot.org/blue/examples.shtml

Thanks
NItin

Nitin when u divide 2222^n/7=3^n(here n is 5555)
Same applies to 5555^2222/7=4^2222
Now hoping u know fermat's theorum,3^6/7=1 & 4^6/7=1
So 3^(6*925)*3^5/7=3^5/7
So, what is left 3^5/7=5 & 4^2/7
Therefore total rem=(5+2)/7=0
Hope u got it

[mathsfreak]

Originally Posted by vikram_k51

These r the questions frm AimCat0718:

1.Remainder when 90^91 is divided by 13?

a.0 b.7 c.12 d.1

2.Last digit of the LCM of(3^2003-1) and (3^2003+1) :

a.8 b.2 c.4 d.6


For 1st one the 91=13*7=n*(13)

So by Fermat's Little theorem is the remainder not 1?Since 2 and 13 are co-primes.:neutral:

hi vikram
1) answer is 12

90^91mod13 = (-1)^91mod13 = -1mod13 = 12

2) answer is 4

Lcm = (3^4006-1)/2
for the last digit,
(3^4006-1)/2mod10
3^4006mod10 = 9 ( euler's )
so 9-1/2
= 4

plz chk and let me know the correct answers.

[vikram_k51]

Originally Posted by mathsfreak

hi vikram
1) answer is 12

90^91mod13 = (-1)^91mod13 = -1mod13 = 12

2) answer is 4

Lcm = (3^4006-1)/2
for the last digit,
(3^4006-1)/2mod10
3^4006
mod10 = 9 ( euler's )
so 9-1/2
= 4

plz chk and let me know the correct answers.

Bingo dude. Both the answers r perfect.

Got it dude,but tell me sthng.Is this the fermat's rule u r using?
Cos as far as I remember according to that if two nos. x,y are co-prime and the dividend is of the form x^n(p) where p is the divisor then x^n( p)mod p=1

And for the 2nd one 4006=445*9+1
Since 9 is the Euler Number for 10 thus by Euler's theorem it is of the form 3*3^(n-1) where n is the Euler Number 9.Then is the remainder here not 3?Thereby is it not of the form(3*1-1)/2=1?

Dude these r very basic doubts and mght seem flimsy but do help me wid them.They cost me nearly 10 mins. and abt 10 marks in the test.:(

[kingnitin]

Originally Posted by ashish banaya..

Nitin when u divide 2222^n/7=3^n(here n is 5555)
Same applies to 5555^2222/7=4^2222
Now hoping u know fermat's theorum,3^6/7=1 & 4^6/7=1
So 3^(6*925)*3^5/7=3^5/7
So, what is left 3^5/7=5 & 4^2/7
Therefore total rem=(5+2)/7=0
Hope u got it

Thanks ashish

But I am still wondering how did u get 3^5555 as remainder:
2222^n/7=3^n(here n is 5555)

Did you actually perform the division or there is some smart mechanism? It is actually this point that is troubling me?

Similarly for this one: 3^5/7 as remainder from [3^(6*925)X3^5]/7=3^5/7

If you can explain this a bit more I think my problem will be solved

[mathsfreak]

@ vikram
i applied remainder theorem in the first one.
90^91mod13
apply remainder theorem here.
v get 90mod13 = 12 = -1
fermet's theorm states a^n(p-1)modp = 1


for the 2nd one, v have to find the remainder of 3^4006mod10
euler of 10 is 4. divide 4006 by 4, the reaminder is 2. hence it is reduced to
3^2-1 = 8
since v have to divide it by 2, the remainder is 4.

@nitin
ashish has applied remainder theorem here.
2222^n/7
divide 2222 by 7, the remainder is 3, hence it is equal to 3^n

hope its clear.