(5 on base 8) multiplied by (6 on base 8) = (36 on base 8)
(5 on base 9) multiplied by (6 on base 9) = (33 on base 8)
(5 on base 10) multiplied by (6 on base 10) = (30 on base 8)
Converting everything into the decimal notation, we get:
5 x 6 = 3N + A
or, 3N = 30 – A
or, N = (30 – A)/3 ---------- (1)
Now, A being the digit of a number, the possible values that A can take are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
Also it is clear from equation (1) that A has to be divisible by 3. Hence now the possible values of A are 0, 3, 6 and 9. In that case, the corresponding values for N are 10, 9, 8 and 7 respectively.
But since the base is N and A is the digit of a number written in base N, A has to be less than N. Hence A can take values 0, 3 and 6 only and corresponding to each of these values of A, we can get three different values for N.
Thus there are 3 values of N satisfying the given conditions of the problem.
there is a funda...if 1's are are repeated n-1 no. of times where n is a prime number greater than 5,it is divisble by n.
e.g. 111111(6 times)%7 = 0
111111111111(12 times) % 13 = 0.
working on similar lines....
7777777 ........upto 56 times % 19
=7*(11111 ........upto 56 times)%19
now 11111 ........upto 18 times will be divisible by 19
so will be 11111 ........upto 36 times
n 11111 ........upto 54 times
7*(11111 ........upto 54 times x 100 + 11)%19
7*(11111 ........upto 54 times x 100 will be divisible by 19
so remainder is 77%19 = 1
N can either be 8, 9 and 10
Possible values 3
Hey Devi, iska explanation to batao ?
(5 on base 8) multiplied by (6 on base 8) = (36 on base 8)
(5 on base 9) multiplied by (6 on base 9) = (33 on base 8)
(5 on base 10) multiplied by (6 on base 10) = (30 on base 8)
N = 3
i am also following the same book
i do not think 7 is possible becoz in that case the value of A will be 9 that is not possible
(5)N x (6)N = (3A)N
Converting everything into the decimal notation, we get:
5 x 6 = 3N + A
or, 3N = 30 – A
or, N = (30 – A)/3 ---------- (1)
Now, A being the digit of a number, the possible values that A can take are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
Also it is clear from equation (1) that A has to be divisible by 3. Hence now the possible values of A are 0, 3, 6 and 9. In that case, the corresponding values for N are 10, 9, 8 and 7 respectively.
But since the base is N and A is the digit of a number written in base N, A has to be less than N. Hence A can take values 0, 3 and 6 only and corresponding to each of these values of A, we can get three different values for N.
Thus there are 3 values of N satisfying the given conditions of the problem.
Thank You.
Ravi Raja
n/a
This comment has been moved here.
A no. 2N is having 28 factors and 3N is haing 30 factors. How many factors will 6N have?
(2N is 2 multiplied by N and similarly)
p;z solve this nishit sinha q
plz solve
What is the remainder when 7777777 ........upto 56 times is divided by 19
there is a funda...if 1's are are repeated n-1 no. of times where n is a prime number greater than 5,it is divisble by n.
e.g. 111111(6 times)%7 = 0
111111111111(12 times) % 13 = 0.
working on similar lines....
7777777 ........upto 56 times % 19
=7*(11111 ........upto 56 times)%19
now 11111 ........upto 18 times will be divisible by 19
so will be 11111 ........upto 36 times
n 11111 ........upto 54 times
7*(11111 ........upto 54 times x 100 + 11)%19
7*(11111 ........upto 54 times x 100 will be divisible by 19
so remainder is 77%19 = 1
reach me at shaheen.csc@gmail.com if u hv any doubts......
regards,
shaheen
Reminder is 13 ! ? plz let me know or it is 4?!
n/a
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