Number System - Questions & Discussions

MBA CAT 2022
1714
hey puys, plz solve dis 4 me...!
Q.Find the remainder when 1212..1212(300 digits) is divided by 999 ?


We know that 1000 leaves a remainder of 1 when divided by 999.
=> 10^3n will leave a remainder of 1 when divided by 999.

12121212....(300 digits) = 121*1000^99 + 212*1000^98 + 121*1000^97 + ..... + 212
= 121 + 212 + 121 + .. + 212 (mod 999)
= 50(121 + 212) (mod 999)
= 50*333 (mod 999)
= 16650 (mod 999)
= 666 (mod 999)
1715

Starting from 5,7... will take a lil bit of time....
We can just divided 500 by 45 and see that the quotient will be 11...
therefore A=11

1716
hey puys, plz solve dis 4 me...!
Q.Find the remainder when 1212..1212(300 digits) is divided by 999 ?


Hi

There is a method to find divisibility by (10^n+1) and (10^n-1), the method depends on value of n and sign.

Here n=3 and sign=-ve.
So we will take three digits at a time, and add them, and find the remainder of the resulting number by 999.

In our case number, x = 1212....121212(300 digits)
so we find
sum = 121 + 212 + 121 + 212... 100 terms
=> sum = (121+121+... 50 terms) + (212+212+212+... 50 terms)
=> sum = (121*50) + (212*50) = (333*50) = 16650
=> remainder with 999 = 666.

Hence your answer = 666.

Varun.
1717

thanx guys.....
here is one more problem :-
Let N=4444^3333. The digits of N are added and this process is repeated until a single digit value K is obtained. Find the value of K ??

1718

how many numbers less than 100000 have the sum of the digits equal to 10?
1. 1381
2.1001
3. 996
4. 252

1719
thanx guys.....
here is one more problem :-
Let N=4444^3333. The digits of N are added and this process is repeated until a single digit value K is obtained. Find the value of K ??


Find the remainder when the given number will be divided by 9, that will be the required answer.

=> 4444^3333 = -2^3333(mod 9) = -8^1111 (mod 9) = 1 (mod 9)

=> Answer would be 1.

how many numbers less than 100000 have the sum of the digits equal to 10?
1. 1381
2.1001
3. 996
4. 252


Let the number be abcde
a + b + c + d + e = 10
Number of solutions to this eq = C(14, 10) = 1001

But we need to subtract those cases when a or b or c or d or e are zero, i.e, 5 cases

=> Required number of numbers = 1001 - 5 = 996
1720
Find the remainder when the given number will be divided by 9, that will be the required answer.

=> 4444^3333 = -2^3333(mod 9) = -8^1111 (mod 9) = 1 (mod 9)

=> Answer would be 9.


how do we know to divide the number by 9...?
nd also the remainder is 1 and how u relate this with the answer 9??

PLZ elaborate....!!!
1721
how do we know to divide the number by 9...?
nd also the remainder is 1 and how u relate this with the answer 9??

PLZ elaborate....!!!


Sorry, answer would be 1, that was a typo.

Say, a1a2a3a4..an is a 'n' digit number, then
a1a2a3..an = (a1 + a2 + a3 + .. + an) (mod 9)
=> If a1a2...an is of form 9K + a, then Sum of digits will also be of form 9k + a.

I hope it helps
1722

hi puys,
n thnx chillfactor...



plz look thru d next problem:-

Q:-A rectangle is such that it can be perfectly cut into smaller squares of a maximum possible side of length 12 units. It is also known that the perimeter of such a rectangle is 384units. How many distinct rectangles satisfy the criteria given.?

1723

Q:-A rectangle is such that it can be perfectly cut into smaller squares of a maximum possible side of length 12 units. It is also known that the perimeter of such a rectangle is 384units. How many distinct rectangles satisfy the criteria given.?


Since, the rectangle is such that it can be perfectly cut into smaller squares of a maximum possible side of length 12 units, we can say that the two sides will be 12a and 12b, where a and b are coprime.

=> 12(a + b) = 192
=> a + b = 16
Possible values = (1, 15), (3, 13), (5, 11), (7, 9)

So, 4 distinct rectangles satisfy the given criteria.
1724

"Let the number be abcde
a + b + c + d + e = 10
Number of solutions to this eq = C(14, 10) = 1001

But we need to subtract those cases when a or b or c or d or e are zero, i.e, 5 cases

=> Required number of numbers = 1001 - 5 = 996"

only 'a' cannot take the value 0 right? the other variables can take 0..so, why did u subtract 5??

is it because the number could be a 2 digit number, 3, 4 or 5..so that is why 0 should not be 1 of the values the variables can take?

1725
"Let the number be abcde
a + b + c + d + e = 10
Number of solutions to this eq = C(14, 10) = 1001

But we need to subtract those cases when a or b or c or d or e are zero, i.e, 5 cases

=> Required number of numbers = 1001 - 5 = 996"

only 'a' cannot take the value 0 right? the other variables can take 0..so, why did u subtract 5??

is it because the number could be a 2 digit number, 3, 4 or 5..so that is why 0 should not be 1 of the values the variables can take?


a + b + c + d + e = 10
We know that all a, b, c, d, e should be less than 10 (as they are single digit numbers)

a can be zero, then the number will be a 4-digit number
a and b both can be zero, in case the number is a 3-digit number and so on. But a, b, c, d all can not be zero, as then e = 10 which is not possible

So, we need to subtract those cases when a = 10 -> 1 case, b = 10 -> 1 case, c = 10 -> 1 case, d = 10 -> 1 case, e = 10 -> 1 case
1726

Please elaborate on how did u get the no of solutions as = C(14, 10)?

1727

1. Find the remainder when 128^1000 is divided by 153.

2. Find the remainder when 50^51^52 is divided by 11.

1728
1. Find the remainder when 128^1000 is divided by 153.

2. Find the remainder when 50^51^52 is divided by 11.


For first one ans is 1
For second one ans is 6
1729
For first one ans is 1
For second one ans is 6


oa is..
1. 51
2. 6

my ans for 1st ques is same as urs but ans is given 51 in book....
Also post your detailed approach for 2nd quest.
1730

hi puys,

V is a 56 digit number. All the digits except the 32nd from the right are the same. If V is divisible by 13, then which of the following can never be the units digit of V?

a.4
b.7
c.1
d.4 or 7 both

1731

plz help me with this
(12)base n+(34)base n+....+() base n, where N is the base of number system, N=2M+1 and M>2. Express X+M+2 in base M

1732
hi puys,

V is a 56 digit number. All the digits except the 32nd from the right are the same. If V is divisible by 13, then which of the following can never be the units digit of V?

a.4
b.7
c.1
d.4 or 7 both

is the answer option d??
1733
aditya131025 Says
is the answer option d??

yes answer is option d.