both are prime number. what else to check to tell them co-prime.
using Euler theorem
Euler no if 59 =58.
61^58*61/59= 61/59=> remainder is 2.
Its not euler theorem , its called Fermat's little theorem. Anyway perfect way to solve.
Its not euler theorem , its called Fermat's little theorem. Anyway perfect way to solve.
Find the last non zero digit of 34! 
All the best
All the best
cyborg5021a SaysIts not euler theorem , its called Fermat's little theorem. Anyway perfect way to solve.
have a look at this link
http://www.pagalguy.com/discussions/ahmedabad-study-group-for-cat-2010-25041201
cyborg5021a SaysIts not euler theorem , its called Fermat's little theorem. Anyway perfect way to solve.
Fermat's little theorem is just a specific case of Euler theorem
For your info....
Find the last non zero digit of 34!
All the best
Its 2...For All reasons
Last Non-Zero Digit For 30! = 4^3 * = 8
34! = 30!*31*32*33*34 = 8*2*3*4 = 2
Its 2...For All reasons
Last Non-Zero Digit For 30! = 4^3 * = 8
34! = 30!*31*32*33*34 = 8*2*3*4 = 2
kindly explain it friend:splat:
imlavmishra Sayskindly explain it friend:splat:
to find last non zero digit of any factorial in format of 10X. ( 10!,20!,30!....)
formula is
4^n*(2*n)!. where n = N/10. last digit of this will give you the answer.
suppose we want to find last non zero digit of
20!. here n=20/10=2.
=>4^2*(2*2)!= 16*24 last digit is 4 so answer is 4.
How many two-digit numbers increase by 18 when their digits are reversed?
(1) 5 (2) 6 (3) 7 (4) 8 (4) 10
every consecutive 2 digit number will give us diff of 9...i;e 54 n reverse 45 will give us 9..so to have a diff of 18 we have to take numbers in dat got diff of 2
13
24
35
46
57
68
so total 7 numbers
79
tell em how u solv this ques.....
jaigurudevshri Saystell em how u solv this ques.....
please elaborate further....you suggesting to what???which post??
jaigurudevshri Saystell em how u solv this ques.....
drkunjan Saysplease elaborate further....you suggesting to what???which post??
Which question??? Pls use quote
Find the last non zero digit of 34!
All the best
is last digit 2...
:
viveknitw Says61^59 % 59 ?
is it 2
How many two-digit numbers increase by 18 when their digits are reversed?
(1) 5 (2) 6 (3) 7 (4) 8 (4) 10
hope is good thing,maybe the best of things.And no good thing ever dies.
- Andy Dufresne.
hi drunkjan ......
it's 7 ......
How many two-digit numbers increase by 18 when their digits are reversed?
(1) 5 (2) 6 (3) 7 (4) 8 (4) 10
my take :
option 3) 7
61^59 % 59 ?
Its 2 !!!!
Its 2 !!!!
viveknitw Says61^59 % 59 ?
remainder = 2
guys just a few questions i am posting pls help me !!!!
Q.1 Find the remainder when 6666.....2002 times is divided by 2002 ?
(1) 663 (2) 660 (3) 0 (4) 66 (5) None of these
Q.2 Find the remainder when 1^17+2^17+3^17+......100^17 is divided by 9 ?
(1) 8 (2) 3 (3) 2 (4) 0 (5) None of these
Q.3 Find the last two digits of the number 3^15^17^19^21
(1) 01 (2) 07 (3) 17 (4) 41 (5) 81
guys just a few questions i am posting pls help me !!!!
Q.1 Find the remainder when 6666.....2002 times is divided by 2002 ?
(1) 663 (2) 660 (3) 0 (4) 66 (5) None of these
Q.2 Find the remainder when 1^17+2^17+3^17+......100^17 is divided by 9 ?
(1) 8 (2) 3 (3) 2 (4) 0 (5) None of these
Q.3 Find the last two digits of the number 3^15^17^19^21
(1) 01 (2) 07 (3) 17 (4) 41 (5) 81
1.
option 2) 660
3333......2002 mod 7 =
3333 mod 7 = 1
333.....2002 mod 11 = 0
3333....2002 mod 13 =
3333 mod 13 = 5
11k = 7p+1 = 13q+5
remainder = 330 *2 = 660
2.
option 5) NOT, remainder = 1
1^17+2^17+3^17+......100^17 mod 9 =
1^5+2^5+3^5+......100^5 mod 9 =
(1+5+0+7+2+0+4+8+0)*11 + 1 mod 9 = 1
3.
option 2) 07
3^15^17^19^21 ==>3^15^odd ==> 3^xx75 ==> 27*xx41 = xx07
hey i dnt need the ans bro i know them...pls tell me the method to solve them
!!!!!