Number System

[naga25french]

Originally Posted by Insanebot

Dude r da options correct?

options are indeed right.. check ma above post for solution..

[prade]

Originally Posted by aniruddha_recd

Hi puys,

Please solve this problem and post the approach too

Q.What will be the remainder when 12121212....300 times is divided by 99?

a.18 b.81 c.54 d. 36

12121212......................300 times
= 121212................00 + 12
= 121212................ times * (99 + 1) + 12
hence 121212...........300 times mod 99 = 12*150 mod 99 = 18 mod 99

[aniruddha_recd]

Originally Posted by prade

12121212......................300 times
= 121212................00 + 12
= 121212................ times * (99 + 1) + 12
hence 121212...........300 times mod 99 = 12*150 mod 99 = 18 mod 99

The answer is correct.Can you please explain how you arrived at the line in bold from the previous step

TIA

[naga25french]

Originally Posted by aniruddha_recd

The answer is correct.Can you please explain how you arrived at the line in bold from the previous step

TIA

see ma post previous page.. i guess that method is relatively easier than this...

[aniruddha_recd]

Originally Posted by naga25french

12121212....300 times

shorcut for number divided by 99 :

take two digits from lefthand side of the number and then add it with next two numbers and carry on till u reach right most number..
divide that number by 99...

by above shortcut,
so we get

12 + 12 + 12( upto 150 times)

12 * 150 = 1800...

1800/99 , the remainder is obviously 18

answer is a) 18..


hope its clear..

Just a small query,if the number of digits of the number is odd,should we count from left?

[prade]

Originally Posted by aniruddha_recd

Just a small query,if the number of digits of the number is odd,should we count from left?

Lets us consider the following case
C=121212121212......12121... 301 digits
= 121212........................120.. 301 digits + 1
= 1212.............................12. 300digits*10 + 1
now C mod 99
= [(1212...........................1200 +12)*10 + 1] mod 99
= [(1212...........................12(99+1) +12)*10 + 1] mod 99

(1212...........................12(99+1) +12) mod 99 = 12*150 mod 99 = 18 mod 99
hence C mod 99 = (18*10 + 1) mod 99 = 82 mod 99

[Insanebot]

Originally Posted by prade

Lets us consider the following case
C=121212121212......12121... 301 digits
= 121212........................120.. 301 digits + 1
= 1212.............................12. 300digits*10 + 1
now C mod 99
= [(1212...........................1200 +12)*10 + 1] mod 99
= [(1212...........................12(99+1) +12)*10 + 1] mod 99

(1212...........................12(99+1) +12) mod 99 = 12*150 mod 99 = 18 mod 99
hence C mod 99 = (18*10 + 1) mod 99 = 82 mod 99

Pls puys temme wot dis mod method is
pls!

[bhandari4u]

Originally Posted by Insanebot

Pls puys temme wot dis mod method is
pls!

let me have the oppurtunity to answer your question. take a look at this:


find the remainder when 2^21 is divided by 5?


write 2^21 in a a way which is simpler to be divided by 5.

let me write it as:

8^7

Now on dividing 8 by 5 gives 3 as remainder (the so called mod operation) which means that on dividing 8^7 will give me 3^7 as remainder (3 from each of the8's)

let me write 3^7 as:
9*9*9*3

when 9 is divided by 5, the remiander is 4 (the result of mod operation is 4) and hence when (9^3)*3 is divided by 5, the remainder is:

(4^3)*3 = 16*4*3 = 1*12 = 1*2 = 2 (do you notice what operation I am applying here?)

hence the remainder when 2^21 is divided by 5 is 2.


cross check by dividing 2097152 by 5.

The beauty of mod operation is that the better you split your original expression, the lesser the time it takes to find the solution. e.g. if instead I had split 2^21 as
128^3. I would have gone like this:

2^21=128^3=3^3=27=2 (mod operation is being apllied at each step)


hope i have made some base. the more you practice, the better(faster) will be the results.

[FakkadNath]

Originally Posted by bhandari4u

let me have the oppurtunity to answer your question. take a look at this:


find the remainder when 2^21 is divided by 5?


write 2^21 in a a way which is simpler to be divided by 5.

let me write it as:

8^7

Now on dividing.......

taking it a li'l ahead,

2^21 can be written as- 2*2^20

and, 2*2^20 = 2* 4^10

4 divided by 5 will give a remainder of -1 (equivalent to 4)

so, 2* (-1) ^10 = 2 * 1 = 2

so, 2^21 mod 5 = 2

[Fuzon]

Quote:
Originally Posted by bhandari4u (Number System)
let me have the oppurtunity to answer your question. take a look at this:


find the remainder when 2^21 is divided by 5?


write 2^21 in a a way which is simpler to be divided by 5.

let me write it as:

8^7

Now on dividing.......


taking it a li'l ahead,

2^21 can be written as- 2*2^20

and, 2*2^20 = 2* 4^10

4 divided by 5 will give a remainder of -1 (equivalent to 4)

so, 2* (-1) ^10 = 2 * 1 = 2

so, 2^21 mod 5 = 2

Lemme take this a further more ahead

see 2^21 will have its last digit 2 , as power cycle of 2 i 4 and 4*5 + 1= 20

Thus the number 2^21 will end in 2

No matter what the other digits are , if the units digit is 2 , then the remainder when divided by 5 OUGHT to be 2