Two nos a and b are co-primes, with b
Option (c).
Let k is the HCF of b and c, then
b = km
c = kn
a = km*p + kn = k(mp + n)
=> HCF of a and b will be k
=> k = 1
=> b and c are co-prime.
Option (c).
Let k is the HCF of b and c, then
b = km
c = kn
a = km*p + kn = k(mp + n)
=> HCF of a and b will be k
=> k = 1
=> b and c are co-prime.
a=bx+c , so it is obvious that b and c are co primes
prikhamjo Saysa=bx+c , so it is obvious that b and c are co primes
Not necessarily.
Its true only when a and b are co-prime.
For eg:-
20 = 2 (mod 18 )
=> 20 = 18*1 + 2
Here 20 and 18 are not coprime, thats why b and c are not coprime.
But when a and b are coprime, say 19 and 7, then
19 = 7*2 + 5
Here 7 and 5, i.e., b and c are coprime
Option (c).
Let k is the HCF of b and c, then
b = km
c = kn
a = km*p + kn = k(mp + n)
=> HCF of a and b will be k
=> k = 1
=> b and c are co-prime.
Its option 2.
Two nos a and b are co-primes, with b
Should be ...
What's the OA..???
rohitaryan Sayswhat is the remainder of 21^199^7/27?
21^199^7/27R = 27*X/27R (As its more than 3 as power of 21, we will get a 27 in the numerator as well) = 0
Remainder is 23
using Chinese Remainder theorem ..
5555...... / ((7^2)*2)
555.... / 2 will yield remainder 1
555..../49 wil yield remainder 16
2k + 1 = 7l + 16
on solving we get l = 1
so remainder is 23
Too good..thanks a ton mate.
remainder is 23
using chinese remainder theorem ..
5555...... / ((7^2)*2)
555.... / 2 will yield remainder 1
555..../49 wil yield remainder 16
2k + 1 = 7l + 16
on solving we get l = 1
so remainder is 23
can this question be solved by any other method ?
Remainder is 23
using Chinese Remainder theorem ..
5555...... / ((7^2)*2)
555.... / 2 will yield remainder 1
555..../49 wil yield remainder 16
2k + 1 = 7l + 16
on solving we get l = 1
so remainder is 23
How did you calculate this -
555..../49 will yield remainder 16
prikhamjo Sayscan this question be solved by any other method ?
98 = 2 * 7^2
So No of 5's taken together which when divided by 98 will give remainder as 0 is 98*(1 - (1/7))*(1 - (1/2))
= 7 * 6 = 42
So we can cancel out 84 '5's from 93 '5's
We are left out with 9 '5's
I dont know any short procedure from this point onwards.
On calculating manually,
555555555/98 gives a remainder of 23.
can anybody pls help me..what will be the last digit of 27 raised to power 36. 😲
mukulsriv10 Sayscan anybody pls help me..what will be the last digit of 27 raised to power 36. :O
1???
mukulsriv10 Sayscan anybody pls help me..what will be the last digit of 27 raised to power 36. :O
36^27 will give last digit as 6.(As, every power of 6 will give the last digit as 6)
27^36 will give the last digit as 1 (Power cycle of 7 is 7, 9, 3, 1).
Two nos a and b are co-primes, with b
I think the answer option will be option 2 here, b and c are co-prime.
Hi Puys,
Here's a new thread specifically for Clocks and Calendars:
http://www.pagalguy.com/forum/prep-r...rs-clocks.html (Questions and tips on Calendars and Clocks) 
mukulsriv10 Sayscan anybody pls help me..what will be the last digit of 27 raised to power 36. :O
27^36. As power is divisible by 4 and last digit of base is odd, last digit will be 1.
pirateiim478 Says27^36. As power is divisible by 4 and last digit of base is odd, last digit will be 1.
Yes to calculate last digit , forget the 10s digit , so 7^36
last digit is 1
pirateiim478 Says27^36. As power is divisible by 4 and last digit of base is odd, last digit will be 1.
27/4R = (-1)
Then, (27^36)/4|R = (27/4R)^36 = (-1)^36 = 1.
Great information!
1)Is the three-digit number X64 a perfect square? options yes, no, cant't say
2)Is the four-digit number PQ36 a perfect square? options yes,no,can't say
3)The number of digits in the square root of a thirteen digit number is? options 6,7,8,either 6 or 7
4)The number of digits in (2PQR)^4 where 2PQR is a four digit number is? options 13,14,15,can't say
5)The number of digits in the cube root of a 29 digit number is? options 10,9,8,can't say
please give justification to the answers