remainder of (7^27)/64 ?
If n=m+1, where m is the product of four consecutive positive integers, then which of the following is /are true
(A) n is odd
(B) n is not a multiple of 3
(C) n is a perfect square
a) all three b) A and B only c) A and C only d) none of these
kpcat Saysremainder of (7^27)/64 ?
7^3*7^3..9 times /64
23*23*23..9 times/64
17*17*17*23 /64
is the ans 15 whats the oa?
my take-
a) all three
kpcat Saysremainder of (7^27)/64 ?
couldnt think of anything so donkey approach.
7^27%64 = (7^3)^9%64 = 23^9%64 = 23*(23^2)^4%64 = 23*17^2*17^2%64 = 23*61*61%64 = 23*-3*-3%64 = 23*9%64 = 15
If n=m+1, where m is the product of four consecutive positive integers, then which of the following is /are true
(A) n is odd
(B) n is not a multiple of 3
(C) n is a perfect square
a) all three b) A and B only c) A and C only d) none of these
All there?? just hit and trial with 2 3 different values.
My SBI ATM PIN is a four digit number.And my HDFC bank ATM PIN is also a four digit number formed by the same four digits as those in ICICI pin but with a different order.When I subtract the two numbers I get a four digit number with the first three digits as 2,3 and 9. What is the unit digit of the difference?
The answer would be a) All three
The first two options can be easily deduced. And we can say that n will be a perfect square by using hit and trial method.
If n=m+1, where m is the product of four consecutive positive integers, then which of the following is /are true
(A) n is odd
(B) n is not a multiple of 3
(C) n is a perfect square
a) all three b) A and B only c) A and C only d) none of these
all three..
kpcat Saysremainder of (7^27)/64 ?
whats d OA..I m getting 15...
Originally Posted by caartey View Post
The possible values of X are 1,7,11,13,17,19,23,29;
So No. of possible values of X is 8
which can be easily determined by finding the euler totient of 30, which basically represents the number of numbers less than and CO-PRIME to 30
E.T = 30(1-1/2)(1-1/3)(1-1/5) = 8
where 2,3,5 are prime factors of 30
Bro...altho i was able to do this sum my own way....can u elaborate on Euler thing as mentioned by you...may b d formula applicable here....thnks in advance....
CHEERS!!
51^203/7=2^203/7=8^67*2^2%7=4....since we can write 8^67 as (7+1)^67
shrutishrma113 SaysMy SBI ATM PIN is a four digit number.And my HDFC bank ATM PIN is also a four digit number formed by the same four digits as those in ICICI pin but with a different order.When I subtract the two numbers I get a four digit number with the first three digits as 2,3 and 9. What is the unit digit of the difference?
Is it 4? If yes, would post the approach
couldnt think of anything so donkey approach.
7^27%64 = (7^3)^9%64 = 23^9%64 = 23*(23^2)^4%64 = 23*17^2*17^2%64 = 23*61*61%64 = 23*-3*-3%64 = 23*9%64 = 15
7^27%64 = (7^3)^9%64 = 23^9%64 = 23*(23^2)^4%64 = 23*17^2*17^2%64 =23*33*33%64=23*1089%64=23*1%64=23
hey i m getting the above as answer . plz tell whether i am making some mistake or what .
7^27%64 = (7^3)^9%64 = 23^9%64 = 23*(23^2)^4%64 = 23*17^2*17^2%64 =23*33*33%64=23*1089%64=23*1%64=23
hey i m getting the above as answer . plz tell whether i am making some mistake or what .
I am also getting 23.
7^27%64 = (7^3)^9%64 = 23^9%64 = 23*(23^2)^4%64 = 23*17^2*17^2%64 =23*33*33%64=23*1089%64=23*1%64=23
hey i m getting the above as answer . plz tell whether i am making some mistake or what .
coolsur SaysI am also getting 23.
yeah mea culpa. 23 it is. calc mistake. hence the notion, avoid donkey apprch.
64 = 4*16
E(4) = 2
27%2 =1
=>7^1%4 = 3
E(16)=8
27%8=3
=>7^3%16=7
hence rem is of the form 16k+7=4k+3
which is 23.
yeah mea culpa. 23 it is. calc mistake. hence the notion, avoid donkey apprch.
64 = 4*16
E(4) = 2
27%2 =1
=>7^1%4 = 3
E(16)=8
27%8=3
=>7^3%16=7
hence rem is of the form 16k+7=4k+3
which is 23.
i had thought on similar lines . but then the 'donkey' approach is a sure shot method . .
Atul Mishra Saysi had thought on similar lines . but then the 'donkey' approach is a sure shot method . .
yeah, but u tend to make calc mistakes, i mean atleast i did.
plus the approach doesnt matter, the end result does. kaise bi le aayo. chahe tukka maara ho.
N= 888888888.....
total 89 digits in N
what is the reaminder when 'N' is divided by 470?
N= 888888888.....
total 89 digits in N
what is the reaminder when 'N' is divided by 470?
470 = 2*5*47
the number is completely div by 2.
it wld leave a remainder of 3 with 5.
and with 47, 4 is what is leaves.
so rem is of the form 47k+4=5k+3=2k
98 fits.
470 = 2*5*47
the number is completely div by 2.
it wld leave a remainder of 3 with 5.
and with 47, 4 is what is leaves.
so rem is of the form 47k+4=5k+3=2k
98 fits.
How did you get remainder as 4 when N is divided by 47? Would you plese explain it in detail?
Paritosh1003 SaysHow did you get remainder as 4 when N is divided by 47? Would you plese explain it in detail?
k lets see,
we know that any number repeated (p-1) times is divisible by p. where p is a prime.
so 888888...... till 92 digits will be completely div by 47
in the question we have 88(upto 89 digits)
so x888 mod 47 = 0
since last digit is 8 => 4*47 = 188
last two digit is taken care of
this means x8 leaves remainder 1 when divided by 47..
48 mod 47 = 1
so x = 4 ...
remainder is 4..
hope its clear.