[abhay15783]

Originally Posted by ankur123xyz

last two digits of 7^2008?

Cat 2008 question Easy because 7 is cyclic for last two digits. Else could be solved by following method Euler number for 100=40 Finding the last two digit means finding the remainder by dividing by 100 Hence 7^2008=7^2000*7^8 By euler theorem7^2000/100 gives remainder 1 and hence the question reduces to 7^8/100 Which will be 01 Though the easier method would have been to solve it by cyclicity this method was good when for other numbers are given where 2 digit cyclicity rules don't exist.

[adi1489]

Originally Posted by ankur123xyz

last two digits of 7^2008?

hii..
To get last 2 digits divide it by 100 and find the remainder...By applying euler theorem:
euler of 100 is 40.
so 2008 mod(40)=8.
so 7^2008 can be written as( 7^8 )mod 100
=>(49)^4=01.

[ArjunTheWinner]

What is euler number of a number ? Can u plz explain a little or giv sm pointers?

[MaskedMenace]

Originally Posted by ankur123xyz

can u plz clarify the mod13*7! part..

2(8!)-21(6!)

16*7! - 3*7! = 13*7!

was this ur doubt ?

[MaskedMenace]

Originally Posted by ArjunTheWinner

What is euler number of a number ? Can u plz explain a little or giv sm pointers?

N = a^p*b^q*c^r .... where a,b,c are primes

then Euler number of a number is

N*(1-1/a)(1-1/b)(1-1/c) ...

Eg: E[15] is 8

15 = 5*3

15*(1-1/5)(1-1/3) = 8

[ArjunTheWinner]

Hey!! thanks for this.
Is this concept there in Arun Sharma ?

[target_iima]

N = 11 × 22 × 36 × 412 × 520 ..... 25 terms. Find the highest power of 75 that can divide N.

pls provide explanation..i just stuckedd

[subu_222412]

Originally Posted by subu_222412

Hello guys...please solve this..

What annual payment will discharge a debt of Rs 808 due in 2 years at 2% per
annum ?

1) Rs.200
2) Rs.300
3) Rs.400
4) Rs.350

Hello guys...please help me out with this prob

[@nee!]

Originally Posted by target_iima

N = 11 × 22 × 36 × 412 × 520 ..... 25 terms. Find the highest power of 75 that can divide N.

pls provide explanation..i just stuckedd

My Take


11 --> 1 1*1
22 --> 2 2*1
36 --> 3 3*2
412 -> 4 4*3
520 -> 5 5*4
and so on ..

11,22,36,412,520,630,742,856,972,1090,11110,12132,13156,14182,15210,16240,17272,18306,19342,20380,21420,22462,23506,24552,25600


now we find there are nine 5's

520,630,1090,11110,15210,16240,20380,21420,25600(2 in number) = 10 5's
Number of 3rs : more than 9 !!


75 --> 5 * 5 * 3 So Max Power 75 ^ 5 ..!! Please corect me if i am wrong ..!!

was in a hurry ..!!

[pinaki.majumder]

Originally Posted by MaskedMenace

N = a^p*b^q*c^r .... where a,b,c are primes

then Euler number of a number is

N*(1-1/a)(1-1/b)(1-1/c) ...

Eg: E[15] is 8

15 = 5*3

15*(1-1/5)(1-1/3) = 8

Two small additions:
1. Euler number of a prime number p is always (p-1)

2. For a number x^n / m if we are trying to find the remainder using euler, then x and m must be co-prime. e.g if we try to find remainder of 2^60 / 6 using euler number of 6 it would be wrong because 2 and 6 are not co-primes. So we divide both by 2 to get 2^59 / 3 . Now we can use the Euler number of 3.
[added after late realization]
One important thing here is that the answer from 2^59 / 3 is not the final one. Since we had earlier divided the problem by 2, we have to multiply the result obtained from 2^59 / 3 into 2 to get the final answer. I realized it only after solving a problem send by a fellow puy.