Number System - Questions & Discussions

MBA CAT 2022
281
One from my side:

Q)In an integer d1d2...dk from left satisfy didi+1 for i even. How many integers from 1000 to 9999 have four distinct digits?

rupam_1987 Says
Is it option b) 778 ?

No. thats not the answer. Its 882.
282

Hello Friends,,,

Help me in solving this problem :

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

Didn't understood the answer in totalgadha!!!

283

hello Friends,, Help me in solving this problem !!!!

1: 2^1990 mod 1990 ( Remainder )
.

284
hello Friends,, Help me in solving this problem !!!!

1: 2^1990 mod 1990 ( Remainder )
.

2^1990 mod 199 = 29
2^1990 mod 5 = 4.
So,199a + 29 = 5b+4 = 2c.
On solving we get 2^1990 mod 1990 = 1024.
285

How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?

286
msabhi Says
How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?

Pls provide options....
287
Hello Friends,,,

Help me in solving this problem :

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

Didn't understood the answer in totalgadha!!!


It should leave a remainder of 23 when divided by 98....
288
msabhi Says
How many 7-digit numbers exist which are divisible by 9 and whose last but one digit is 5?


Is the answer 100000?
289
R @ J Says
Is the answer 100000?

R @ J Says
It should leave a remainder of 23 when divided by 98....

Pls provide the solutions.
290
jain_ashu Says
Pls provide the solutions.

I found it in a material which had only this question...answer not given puys :(...am sorry...i cudnt solve it ..so wanted to know aproach and come to conclusion ,which is nothing but puys answer ...my apologies 😞
291
hello Friends,, Help me in solving this problem !!!!

1: 2^1990 mod 1990 ( Remainder )
.

2^1990 mod 199 = 29
2^1990 mod 5 = 4.
So,199a + 29 = 5b+4 = 2c.
On solving we get 2^1990 mod 1990 = 1507.


Ashu, On solving 199a + 29 = 5b+4 = 2c, I'm getting a=5, b=204, c=512, how did u arrive 1507? I thought the ans is 1024..
292
Sahana Kavitha Says
Ashu, On solving 199a + 29 = 5b+4 = 2c, I'm getting a=5, b=204, c=512, how did u arrive 1507? I thought the ans is 1024..

199a + 29 = 5b+4. So, the general case becomes 995m+1024.
Now, 995m+1024 = 2c. at m = 2, c = 1507.
2c = 3014. Reminder is 1024
P.S : Have just applied the chinese reminder theorem
293
Hello Friends,,,

Help me in solving this problem :

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

Didn't understood the answer in totalgadha!!!


= 5(10^92+10^91+10^90.....10^0) % 98
= 5(2^46+ 2^45(10+1)+ 2^44(10+1)+ 2^43(10+1)......2^0(10+1)) % 98
= 5(2^46+ 11(2^45 + 2^44 + 2^43....+2^0)) % 98
= 5(2^46 + 11(2^46 - 1)) % 98
= 5(12*2^46 -11)) % 98
= (60*2^46 - 55) % 98, solving this
= (78-55) % 98
= 23
294
Hello Friends,,,

Help me in solving this problem :

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

Didn't understood the answer in totalgadha!!!

98 = 2*49
now 5555.....93 times mod49 = 9
55555...93timesmod 2 = 1
so the num is of the form 2k+1 = 49m+9
it comes to be 98k+ 107
so the remainder is 107
295
98 = 2*49
now 5555.....93 times mod49 = 9
55555...93timesmod 2 = 1
so the num is of the form 2k+1 = 49m+9
it comes to be 98k+ 107
so the remainder is 107

Pls specify what method have you used for 5555.....93 times mod49 bcoz 49 is not a prime number???
296
= 5(10^92+10^91+10^90.....10^0) % 98
= 5(2^46+ 2^45(10+1)+ 2^44(10+1)+ 2^43(10+1)......2^0(10+1)) % 98
= 5(2^46+ 11(2^45 + 2^44 + 2^43....+2^0)) % 98
= 5(2^46 + 11(2^46 - 1)) % 98
= 5(12*2^46 -11)) % 98
= (60*2^46 - 55) % 98, solving this
= (78-55) % 98
= 23

Could you pls elaborate on the red part? How have you converted powers of 10 to powers of 2 and where have the 5s gone?
Thanx.
297
jain_ashu Says
Pls specify what method have you used for 5555.....93 times mod49 bcoz 49 is not a prime number???

yea srry u r rt...my method is wrong..was trying to apply fermi little theorem...
298
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


Thanks Ashu, totally missed that....:banghead:
299
Could you pls elaborate on the red part? How have you converted powers of 10 to powers of 2 and where have the 5s gone?
Thanx.

= 5(10^92+10^91+10^90.....10^0) % 98
= 5(2^46+ 2^45(10+1)+ 2^44(10+1)+ 2^43(10+1)......2^0(10+1)) % 98
= 5(2^46+ 11(2^45 + 2^44 + 2^43....+2^0)) % 98
= 5(2^46 + 11(2^46 - 1)) % 98
= 5(12*2^46 -11)) % 98
= (60*2^46 - 55) % 98, solving this
= (78-55) % 98
= 23


(10^92) % 98 = (2^46) %98, as every 10^2 % 98 yeilds 2 as remainder, applying remainder theorem

(10^91+10^90) % 98 = (2^45(10+1))%98, as every 10^2 % 98 yeilds 2 as remainder, applying remainder theorem and taking 2^45 common for both the powers (91 and 90).. likewise for all the other term..

Hope this is clear now.
300
hello Friends,, Help me in solving this problem !!!!

1: 2^1990 mod 1990 ( Remainder )
.


My take is:
2^1990%1990 =2^1990%(10*199)
part1:
2^1990%10=2^1990%(2*5) = 2^1989%5= 2^1%5 = 2 *2 =4
The no is 10b+4 (say).

part2:
2^1990%199 = 2^10%199 = 29

So 199a+29 = 10v+4
a=5, b=102
So the remainder is 1024 Ans.