Number System - Questions & Discussions

MBA CAT 2022
321
msabhi Says
How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?



my ans is coming out around 27000 is the ans right or not???
322


The above is wilson's theorem. Can someone explain this to me in simpler terms with an example..
323


The above is wilson's theorem. Can someone explain this to me in simpler terms with an example..


This theorem says like this way that if in case n is a prime no. then when (n-1)! is divided by n then it leaves the remainder -1..

for example: let n be 7
then (7-1)! when divided by 7 will give the remainder -1 which is equivalent to 6..

and similarly with other prime no.

this was the case when n was a prime no..,, but if n is a composite no. then it will leave the remainder 0..
condition n should be greater than 5..(n>5)
324

Originally Posted by orangejawan View Post
Hello Friends,,,

Help me in solving this problem :

remainder of 5555......93times divided by 98

Didn't understood the answer in totalgadha!!!

N = 5555......93times

98 = 2*49

N/2 = 1

N/49 = 9

so 2a + 1 = 49b + 9

so final remainder is 107


Pls explain how do u get a remainder 107, when the a number
N (5555...93 times) is divided by 98? (107>9 am I missing some thing here.. also check the below post and give me your observation

http://www.pagalguy.com/forum/quantitative-questions-and-answers/43728-number-system-ii-30.html#post1652827
326
rohan101 Says
what is the highest power of 3 in 58!-13!...arun sharma ques please explain the method as well


The highest power of 3 in (58!-13!) is the power of 3 in the common term, which is 13!(58*57*....14 -1), hence 13/3+13/9 = 5
327
Originally Posted by orangejawan View Post
Hello Friends,,,

Help me in solving this problem :

remainder of 5555......93times divided by 98

Didn't understood the answer in totalgadha!!!


Pls explain how do u get a remainder 107, when the a number
N (5555...93 times) is divided by 98? (107>9 am I missing some thing here.. also check the below post and give me your observation

http://www.pagalguy.com/forum/quantitative-questions-and-answers/43728-number-system-ii-30.html#post1652827


itz answer can't be 107...
my ans coming out is 23...
328
Originally Posted by orangejawan View Post
Hello Friends,,,

Help me in solving this problem :

remainder of 5555......93times divided by 98

Didn't understood the answer in totalgadha!!!


Pls explain how do u get a remainder 107, when the a number
N (5555...93 times) is divided by 98? (107>9 am I missing some thing here.. also check the below post and give me your observation

http://www.pagalguy.com/forum/quantitative-questions-and-answers/43728-number-system-ii-30.html#post1652827


itz answer can't be 107...
my ans coming out is 23...
329
itz answer can't be 107...
my ans coming out is 23...



sorry for the double post the correct ans is also 23....:-P:-P:-P:-P:-P
330

puys this was a very old post, we didnt find the answer yet, pls try

Originally Posted by v!vek@rora View Post
Divisors & sum of divisors of 12^33+34^23+2^47

SK

331
N=888888888...89 digits
470 = 2 * 5 * 47
N/2 , the remainder is 0
N/5 , the remainder is 3
N/47 , the remainder is 4
2a = 5b + 3 = 47c + 4
10k + 8 = 47c + 4
hence the final remainder is (47 * 2) + 4 = 98


N/47 = 4 HOW? Is there any way to calculate it?
332
Originally Posted by orangejawan View Post
Hello Friends,,,

Help me in solving this problem :

remainder of 5555......93times divided by 98

Didn't understood the answer in totalgadha!!!


Pls explain how do u get a remainder 107, when the a number
N (5555...93 times) is divided by 98? (107>9 am I missing some thing here.. also check the below post and give me your observation

http://www.pagalguy.com/forum/quantitative-questions-and-answers/43728-number-system-ii-30.html#post1652827




ma mistake : answer is 23:banghead::banghead:

yea i realised the flaw .. infact i was the person who corrected ashu in that post .. now i did the same mistake
333
tac007 Says
N/47 = 4 HOW? Is there any way to calculate it?



check this link.. have already given explanation

http://www.pagalguy.com/forum/quantitative-questions-and-answers/43728-number-system-ii-24.html#post1649934
334
msabhi Says
How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?




ten's digit is 5

so now other six digits can be

a+ b+ c+ d+ e + f = 4 or 13 or 22 or 31 or 40 or 49


find the solutions and then add up.. we can obtain the answer
335
ten's digit is 5

so now other six digits can be

a+ b+ c+ d+ e + f = 4 or 13 or 22 or 31 or 40 or 49


find the solutions and then add up.. we can obtain the answer


Agree that this is the way to go but I think its not that easy to calculate.
For a+ b+ c+ d+ e + f = 4.
So, the numbers will be 6!*9C5 - 5!*8C4.
For a+ b+ c+ d+ e + f = 13, 6!*18C5 - 5!*17C4 and so on. It would be very lengthy.
336
msabhi Says
How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?

find the coefficient of x^9 in (1+x+x^2....x^9)^6*x^5
(1-x^10)^6*(1-x)^6*x^5
(1-6x^10)*(1-x)^6*x^5
(1-x)^6*x^5 - 6*(1-x)^6*x^15
4+6-1C5 = 9C5 = 126

This is my first attempt using the above concept...kindly guide me in this, if its wrong thn pls tell me where i m goin wrng....
Thankss in advance
337
find the coefficient of x^9 in (1+x+x^2....x^9)^6*x^5
(1-x^10)^6*(1-x)^6*x^5
(1-6x^10)*(1-x)^6*x^5
(1-x)^6*x^5 - 6*(1-x)^6*x^15
4+6-1C5 = 9C5 = 126

This is my first attempt using the above concept...kindly guide me in this, if its wrong thn pls tell me where i m goin wrng....
Thankss in advance


What is the eqn use have used? I think it is x1+x2+x3+x4+x5+x6 = 4.
In that case it will be coeff of x^4 in (1+x+x^2+...+x^9)^6
= (1-x^10)(1-x)^-6 = (1-x^10)*(1+6C1*x+7C2*x^2+8C3*x^3+9C4*x^4+...)
=9C4 = 126.
But this contains the condition where x1 = 0.
So, use coeff of x^4 in (x+x^2+x^3+...+x^9)*(1+x+x^2+...+x^9)^5 and then find out
338

If (2146!)in base 10 = (x)in base 26, then what will be the number of consecutive zeroes at the end of x?
a. (85)base 10 b. (177)base 10 c. (534)base 10 d. (17base 10 e. None of these

339
What is the eqn use have used? I think it is x1+x2+x3+x4+x5+x6 = 4.
In that case it will be coeff of x^4 in (1+x+x^2+...+x^9)^6
= (1-x^10)(1-x)^-6 = (1-x^10)*(1+6C1*x+7C2*x^2+8C3*x^3+9C4*x^4+...)
=9C4 = 126.
But this contains the condition where x1 = 0.
So, use coeff of x^4 in (x+x^2+x^3+...+x^9)*(1+x+x^2+...+x^9)^5 and then find out

dude i guess we need to find for seven digits...
so i have used x^5 for the second last digit and the multiplied it to the rest of the eqn....wats the overall...and if i m wrong can u solve it for me using the same concept:) thnxx
340
If (2146!)in base 10 = (x)in base 26, then what will be the number of consecutive zeroes at the end of 'x'?
a. (85)base 10 b. (177)base 10 c. (534)base 10 d. (17base 10 e. None of these


is it 85(base10)??
go on dividing 2146! by 26 (to get remainder as 0 which is the total zeros at the end of it in base 26:)
341
is it 85(base10)??
go on dividing 2146! by 26 (to get remainder as 0 which is the total zeros at the end of it in base 26:)

the answer is b. 177(base 10)

even i am not getting this answer.