this is the question: Q. (5 on base N) multiplied by (6 on base N) = (3A on base N), where A is the unit digit of the two digit no. 3A. How many values of N are possible?
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
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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?!
Please re phrase it.
nybody having the password for the online resource of this book?
Plz solve this:
How many number of zeroes will be there at the end of 12! expressed in base 6?
thnaks,
Its like calculating how many 2's and 3's are there in 12! as number of 2's > number of 3's number of zeros at the end = number of 3's [12/3] + [12/9] = 4 + 1 = 5 Thats the answer
Bhai mere we are calcualting number of 0's in base 6 not in base 10. while converting a number from base 10 to base 6, we 'll divide it by 6 if you dont belive me then covert the number 12! to base 6 and chk it
Dear frn why hv u considrd 2 & 3 cz 3 is nt necessary 4 gettin 0.. 10=2 x 5... Drfore 12! consists of [12/5]=2 no. of Zeros in base 6... Giv ur best to d world. Nd d best will cm back to u.... Regards, Dipanjan.....
SORRY YAAR...I hv nt seen dt was in base 6(2 x 3)....So its ok..I hv dn dt mistake..... Regards, Dipanjan....
A number N when divided by 7 gives a remainder of 4. when divided by 11 gives a remainder of 6. whats the remainder when divided by 13. Plz solve and tell me the method. Regards,
See man:
The no. is 7k+4 or 11k1+6
So,
7k+4=11k1+6
=>7k-11k1=2
So,k=5,k1=3
The no. is 39...
As 39=0(mod 13)
Remainder is zero....
Giv ur best to d world.
Nd d best will cm back to u...
Regards,
Dipanjan....
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(a,b) is an ordered pair such that a and b are two digit numbers a is the number obtained by interchanging the digits of b How many ordered pairs are there such that (a-b) is a perfect square greater than Zero? a)3 b)10 c)13 d) none of these
the number of values of N will be 10.
the question simplifies to 3N + A = 30
Now N cannot be 0 and A cannot be negative.
therefore N will have values from 1 to 10 .
correct me if i am wrong
Possible values of N are 8, 9 and 10.
So in total 3
When N=8, (5)b8 * (6)b8 = (36)b8
When N=9, (5)b9 * (6)b9 = (33)b9
When N=10, (5)b10 * (6)b10 = (30)b10
All the answers are of the form 3A