Number System - Questions & Discussions

MBA CAT 2022
301
199a + 29 = 5b+4. So, the general case becomes 995m+1024.
Now, 995m+1024 = 2c. at m = 2, c = 1507.
P.S : Have just applied the chinese reminder theorem



chinese remainder theorem is fine.. but what does it state.. the eqn should be balanced

2a is there .. so definitely remainder should be even number .. so the remainder should be 1024
302
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
303

777^777/1001...remainder???

304
v!vek@rora Says
777^777/1001...remainder???


is it 259 ?
305
v!vek@rora Says
777^777/1001...remainder???


The answer is 259..

1001=13*11*7...
Hence the question shrinks to 777^776 * 111/13 and 777^776 *111/11 with the remainder multiplied with 7 coz we divided with 7 ...

Hence 11a+4=13b+11
This gives a=3,b=2
Hence the remainder is (11a+4)*7=259 😃
306
chinese remainder theorem is fine.. but what does it state.. the eqn should be balanced

2a is there .. so definitely remainder should be even number .. so the remainder should be 1024

My mistake.... I had calculated only c. It should be 2c which is 3014.
So, 3014 - 1990 = 1024.
307
N = 5555......93times

98 = 2*49

N/2 = 1

N/49 = 9

so 2a + 1 = 49b + 9

so final remainder is 107

what method have you employed to find N/49 ???
308
jain_ashu Says
what method have you employed to find N/49 ???


x555 should be divisible by 49 ..

x = 9

so hence remainder = 9
309
x555 should be divisible by 49 ..

x = 9

so hence remainder = 9

Could you explain why x555 should be divisible by 49. Are you saying 5555....96 times is divisible by 49??? but why???
310
x555 should be divisible by 49 ..

x = 9

so hence remainder = 9

but 49 is not a prime then how are u applying fermi little theorem...even i did the same mistake few posts back...
311
Mani_23 Says
ans:sum will be 20*2^6 +10(2^6-1) = 1910


@Mani,
Could u please explain your method? It looks very hady, but how, plz share !!
312
option e)1910
7th position will always hold 1,==> 64*6!/(3!*3!) = 1280
for all other position
*(32+16+8+4+2+1) = 630
total = 1910

Consider all possible seven-digit binary numbers having four 1s and three 0s. What is the sum of all such numbers?

(a) 1470 (b) 1615 (c) 1740 (d) 1825 (e) 1910

Ankur can you tell me how have you calculated the bold part???
313
jain_ashu Says
Ankur can you tell me how have you calculated the bold part???

consider this, 1 will always be there in 7th place
1 _ _ _ _ _ _ , we are left with three1's and three 0's
consider the 6th place:-
1 1 _ _ _ _ _, ways to fill the remaining places = 5!/(3!*2!), for all these numbers 1 will be there in the 6th place
value = 2^5 * (5!/(3!*2!)), a 0 in 6th place will add nothing to the value
5th place, value = 2^4 * (5!/(3!*2!))
similarly for other places.
314
is the ans 24?
My approach...
for the sum of coeficients to be prime the roots shud be even:)
let the quadratic eqn be (x-a)(x-b) = 0
also given that for sum integer the value is -55 = -5 * 11
from the options let the roots be 20 n 4. we can see tht if we subsitute 9 we get (9-4)(9-20) = 5*-11 = -55
hence all conditions are satisfied here...
so the sum of the roots must be 20+4 = 24.
PS-Did a bit of hit n trial before taking the values:)


What is this logic that "for the sum of coeficients to be prime the roots shud be even"??
Plz clarify!!
315
7^m + 7^n
m = 1,3,7,9
n = 1,3,7,9
The probability that it is divisible by 5 is 1/4.


Hey dude can u plz explain it further, because for following pairs of (m,n) the condition satifies:

(m,n)=(2,4),(2,, (4,6),(5,7) so on many more even.
And by the way 1,3,7,9 are not values of m/n. these are unit digits got in powering 7 i think...plz clarify.
316
hey i got this from somewhere(might be from some thread)
see if it hlps
Co-Primes are very important , when it comes to finding remainders.
First things first, What is a Co-Prime ?
2 numbers are said to be co-primes/relatively prime if their HCF is 1.
That is the 2 numbers have no other common factor apart from the number.

Are (1,3) co-prime?
317
hey i got this from somewhere(might be from some thread)
see if it hlps
Co-Primes are very important , when it comes to finding remainders.
First things first, What is a Co-Prime ?
2 numbers are said to be co-primes/relatively prime if their HCF is 1.
That is the 2 numbers have no other common factor apart from the number.

Are (1,3) co-prime?

yea 1 and 3 are coprimes....
7 and 11 are coprimes....basically GCD shud be 1
318
What is this logic that "for the sum of coeficients to be prime the roots shud be even"??
Plz clarify!!

pls click this link here
http://www.pagalguy.com/discussions/number-system-questions-discussions-25043728

check for post no. #162
319
470=10*47
8*10^88+8*10^87+8^86+-------------------------------------8
8(10^89-1)/9 mod 47
8(10^89-1)/9 =k
8*10^89=9k+8
After that I did not clearly, plz clarify after this part===========>>
Why do u multiply by 1000, plz tell.
(8*10^92)mod 47=1000(9k+ mod 47
9000k mod 47 =2
k=43
47 x+43=10y+8
After this ok for me============>>
y=47x+35/10
x=5,y=27
so remainder is 278


plz clarify red part !!
320
pls click this link here
http://www.pagalguy.com/discussions/number-system-questions-discussions-25043728

check for post no. #162

thanx buddy !!