Number System - Questions & Discussions

MBA CAT 2022
924
A is a 4 digit +ve interger and B is a 4 digit +ve integer formed by reversing the digits. A-B is divisible by 45. Find the no.of possible values??

could u help me with this guys...


A = 1000a + 100b + 10c + d
B = 1000d + 100c + 10b + a
A-B = 999(a-d) + 90(b-c)
If A-B Is divided By 45..Then (a-d) Must Be Divisible By 5....And We CAn Have any value of (b-c)
Also a>d
So We have...(a,d) => (6,1),(7,2),(8,3),(9,4)
Total Values => 4*10*10 = 400 Numbers..Correct Me If I Am Wrong
925
A = 1000a + 100b + 10c + d
B = 1000d + 100c + 10b + a
A-B = 999(a-d) + 90(b-c)
If A-B Is divided By 45..Then (a-d) Must Be Divisible By 5....And We CAn Have any value of (b-c)
Also a>d
So We have...(a,d) => (6,1),(7,2),(8,3),(9,4)
Total Values => 4*10*10 = 400 Numbers..Correct Me If I Am Wrong


Dude I have a doubt here .....
Why not a = d ....??

For eg 1431 satisfies the given condition .......

As 1431 - 1341 = 90
926
A = 1000a + 100b + 10c + d
B = 1000d + 100c + 10b + a
A-B = 999(a-d) + 90(b-c)
If A-B Is divided By 45..Then (a-d) Must Be Divisible By 5....And We CAn Have any value of (b-c)
Also a>d
So We have...(a,d) => (6,1),(7,2),(8,3),(9,4)
Total Values => 4*10*10 = 400 Numbers..Correct Me If I Am Wrong


bhai want to correct you as i thnk u r wrong(pj at pg)

jokes apart; y r u considrng a>d yourself
example; A=1006, so B=6001....A-B= -4995
-4995 is divisible by 45(i hav read in divisiblity rules,,,,)
secndly, a-d=0
(1,1),(2,2).....(9,9)

so....17*10*10=1700 values possible

now,correct me if i m wrong
927

the answer for this is 805 values...

928
rahul20001 Says
the answer for this is 805 values...



if we are wrong thn also 805 is not possible......
1st case:a__a=9*10*10=900(whn both first and last are same)
i hope baaki case likhne ki jarurt bhi nahi hai to prove 805 as wrong
929

Q> how many different four digit nos are there in the octal i.e base 8 system, expressed in that system??
a> 3584 b>2058 c>6000 d>7000

guys give the explaination also along with ur ans..

930

Q> how many different four digit nos are there in the octal i.e base 8 system, expressed in that system??
a> 3584 b>2058 c>6000 d>7000

ans is 7000.
guys give the explaination also along with ur ans..

931
if we are wrong thn also 805 is not possible......
1st case:a__a=9*10*10=900(whn both first and last are same)
i hope baaki case likhne ki jarurt bhi nahi hai to prove 805 as wrong


i'll check the answer from the source again and get back....
932
Q> how many different four digit nos are there in the octal i.e base 8 system, expressed in that system??
a> 3584 b>2058 c>6000 d>7000

ans is 7000.
guys give the explaination also along with ur ans..



it is like this that if there is base 8 then there will be 7000 values..
if it is base 7 then there will be 6000 value..
if it is base 6 then 5000 values ,...........and so on for any other base also...

as in base 10 there are 9000 four digit values..
933
I am getting 28.
T can have all elements from 40 to 13. Total = 28.


what if we do not take consequtive no.??
like 1+2+40=43
934
Guys pls give me the detailed solution of these Problems
Q1) What is t remainder when x^99+x^88+x^77+x^66+x^55+x^44+x^33+x^22+x^11+1 is didvided by x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1

Q2) the volume of cuboid is 231cubic centimeter. All dimensions r integers.Find minimum sum of all the cuboid?
a)21 b)63 c)84 d)126
935
Guys pls give me the detailed solution of these Problems
Q1) What is t remainder when x^99+x^88+x^77+x^66+x^55+x^44+x^33+x^22+x^11+1 is didvided by x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1

Q2) the volume of cuboid is 231cubic centimeter. All dimensions r integers.Find minimum sum of all the cuboid?
a)21 b)63 c)84 d)126


ans 1.: conventional method: =x^99+x^88+x^77+x^66+x^55+x^44+x^33+x^22+x^11+1
=(x^9)^11+(x^8 )^11+(x^7)^11+(x^6)^11+(x^5)^11+(x^4)^11+(x^3)^11+(x^2)^11+(x)^11+(1)^11

as we know: a^n + b^n +.........+z^n is divisible by a+b+.....z if n is odd

therfore, remainder is zero.

unconventional: put x=1 n njoy ur life
as 10/10 rem=0

Ans2. r u asking 4 min sum of all the sides:(assuming u r asking 4 same)

231= 3*7*11
3+7+11=21 option (a)
936
Q> how many different four digit nos are there in the octal i.e base 8 system, expressed in that system??
a> 3584 b>2058 c>6000 d>7000

ans is 7000.
guys give the explaination also along with ur ans..



How did u get the answer as 7000.Kindly share your logic or solution provided.

My take => For octal numbers the range of numbers u can use is from 0 to 7.

Thus if "abcd" is your number then "a" can be filled by numbers from 1 to 7 i.e. 7 values.
Similarly for bcd, we get 8*8*8.

Thus in all it comes out to be 7*8*8*8=3584...
937
Q> Let S={1,2,3,,,,,n} be a set of N natural nos. Let T be a subset of S such that the sum of any three elements of T is not less than N.Find the maxi nos of element in any such subset T for N=40??

(a)26
(b)27
(c)28
(d)none of these

The ans is-d

What should be the approach 4 such a question and explain also??

I am getting 28.
T can have all elements from 40 to 13. Total = 28.

what if we do not take consecutive no.??
like 1+2+40=43

Try taking and tell me how many can you take. you won't be able to cross 37. In this case you can take 1,2,37,38,39 and 40. That's it.
938

guys help me out in this ....a number n when divided by a divisor d gives a remainder 52.When 5n is divided by d it gives a remainder of 4.how many values of d are possible?

939

@ajit- is the answer 9?

940

@vivek No it's 3 ..can you post your approach?

941
ajit@mumbai Says
guys help me out in this ....a number n when divided by a divisor d gives a remainder 52.When 5n is divided by d it gives a remainder of 4.how many values of d are possible?


hey ajit,
here is my solution:

N= pd + 52 which shows d is grtr thn 52
5N= 5pd + 260 = qd+4....this implies all the nos grtr thn 52 and factor of 256(260-4) will vallidate this condition....256=2^8....
numbrs possible= 2^(8-6)=2^2=3factors, therefore three nos possible,i.e, 64,128,256
correct me if i m wrong.
942
hey ajit,
here is my solution:

N= pd + 52 which shows d is grtr thn 52
5N= 5pd + 260 = qd+4....this implies all the nos grtr thn 52 and factor of 256(260-4) will vallidate this condition....256=2^8....
numbrs possible= 2^(8-6)=2^2=3factors, therefore three nos possible,i.e, 64,128,256
correct me if i m wrong.



pls explain
943
ajit@mumbai Says
guys help me out in this ....a number n when divided by a divisor d gives a remainder 52.When 5n is divided by d it gives a remainder of 4.how many values of d are possible?


5N = bd + 4..(1)
N = ad + 52..(2)
Multiply 2 by 5
5N = 5ad + 260
Now 260 mod Any Number Must Leave 4....
So Largest Number = 256...
Rest Numbers Are Factors of 256...
So total Factors = 9
But 1,2,4 will not suffice...
Also Value less than 52 will not suffice....
So Values of d = 3