cyborg5021a SaysWhats the remainder when 6^57 is divided by 61?
Reminder is 37.
Reminder is 37.
cyborg5021a SaysWhats the remainder when 6^57 is divided by 61?
2^57 mod 61 = 23
3^57 mod 61 = 52
23*52 mod 61 = 37
thanks a lot
6^ 57/61= 6^-3/61=6^-3*306^3/61 =51^3/61=-10^3/61=-24
thus = 61-24 = 37
m i correct?
plz elucidate from the third step...
i m not able to do after 2nd step coz its becoming tedious to apply inverse euler in that case...
rajiv55bits Saysdude, if u can send me any pdf's regarding quant it'd be helpful. My mail id is , thanks in advance
dont put ur email like this.
if mods c this, it will not be good for u.........
if u want pdfs- u can always browse the attachments in threads!
thanq so much i'm sorry i didn't know that
plz elucidate from the third step...
i m not able to do after 2nd step coz its becoming tedious to apply inverse euler in that case...
i would recommend the method followed by divisht above.
any way ma method is
in the third step i multiplied by 306
Now i need to multiply numerator with a multiple of 6 which leaves remainder 1 with 61
so the cond 6m = 61k+1 => ...which is true for k=5 and the number is 306.
sorry for late reply. 37 is the correct Ans
dont put ur email like this.
if mods c this, it will not be good for u.........
if u want pdfs- u can always browse the attachments in threads!
You did well in suggesting him to remove his email id but your own post contains it. Please edit the quoted portion in your post and remove his email id.
61^59 % 59 ?
viveknitw Says61^59 % 59 ?
61^59 mod 59 =
2^59 mod 59 =
32* (2^6)^9 mod 59 = 32 * 125^3 mod 59 =
32 * 343 mod 59 = 2
remainder = 2
my doubt was - can v use euler here!
even by euler v get the same answer but r 59 and 59 co-prime?
my doubt was - can v use euler here!
even by euler v get the same answer but r 59 and 59 co-prime?
i guess u r mistaking in applying euler. we need to verify if 59 n 61 are coprime.
cyborg5021a Saysi guess u r mistaking in applying euler. we need to verify if 59 n 61 are coprime.
but we ca use Fermat's little theorem here.
59 is prime and 59 and 61 are coprime.
hence 61^58 mod 59 =1
hence 61^59 mod 59 = 2
my doubt was - can v use euler here!
even by euler v get the same answer but r 59 and 59 co-prime?
for two numbers to be co-prime, they should have gcd = 1.
now, decide for yourself.
cyborg5021a Saysi guess u r mistaking in applying euler. we need to verify if 59 n 61 are coprime.
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.
How many two-digit numbers increase by 18 when their digits are reversed?
(1) 5 (2) 6 (3) 7 (4) 8 (4) 10
How many two-digit numbers increase by 18 when their digits are reversed?
(1) 5 (2) 6 (3) 7 (4) 8 (4) 10
numbers of form 11p + 2 satisfies the condition
so 3)7
numbers of form 11p + 2 satisfies the condition
so 3)7
ya right answer but your approach is very nice....
How many two-digit numbers increase by 18 when their digits are reversed?
(1) 5 (2) 6 (3) 7 (4) 8 (4) 10
7 numbers. The difference between the digits should be 2.