Number System - Questions & Discussions

@mbajamesbond said:@Psychamour
1-)Rem[ 3^18/19 ] = 1
2)Rem[ 12^36/37 ] = 1
How come the remainder is 1...where as in 1st taking negative remainder would be -16 and in 2nd negative rem would be -25.....
Please correct me....

Hi @[563226:mbajamesbond]

Im sorry I couldnt get your point. Pls elaborate a little.

For (1) , Since 3 and 19 are coprime, we can use Euler's.

Euler(19) = 18

So, 3 (raised to the power 18) when divided by 19 will give remainder 1.

ie. (3^18)mod19 = 1

Did the 2nd one in the same way.

Euler(37) = 36

I believe neagative remainders are helpful only when they are smaller than the actual number, the number which you are dividing.

May be Im missing something. Pls help me understand your point.

The 150 contestants of Miss World 2012 are given individual numbers from 1 to 150. Several rounds happen before the final winner is selected. The elimination in each round follows an interesting pattern. In the 1st round starting from first contestant, every 3rd contestant is eliminated i.e. 1st, 4th, 7th, .... This repeats again from the first numbered (among the remaining) contestant in the next round (leaving 3, 5, 8, 9, ... ). This process is carried out repeatedly until there is only the winner left. What is the number of Miss World 2012 ?
(a) 93
(b) 48
(c) 119
(d) 38
(e) 140

@[283374:viveknitw]

Ingredients :
Note that in every round 1st, 4th, 7th, 10th..contestant get eliminated.
ie. in every elimination round, (3n+1)th contestant is struck off. 'n' here starts from 0.
********************************************************************
Lets make the dough now :

Start with 150 contestants, read terms, if every (3n+1)th term is to be eliminated, the number of terms eliminated will be 50, since (3.49+1)=148, which is the last term to be struck off.

Initialy number of terms in Round 0) 150
No. of terms left in Round 1) 150 - 50 = 100

Now, for second round we have 100 terms, we again eliminate every (3n+1)th term. Since 100 = 3.33 + 1. The last term will get eliminated in this round. And total number of terms struck off in this round = 33 + 1 = 34, as we start n from 0

So, No. of terms left in Round 2) 100 - 34 = 66

Now for third round we have 66 terms, we again eliminate every (3n+1)th term. Since 66=3.22. This ensures that 66th term will not get eliminated since we are looking for 1 more than the mutliple of 3. So, which term wil get eliminated from the end. The (21.3+1)th or 64th term. And hence the total number of terms eliminated in this round = 21 + 1 = 22

So, No. of terms left in Round 3) 66 - 22 = 44

Continuing in this manner, we obtain the number of terms left after each round as

No. of terms left in Round 4) 44 - (14+1) = 29 ................since 14*3 = 42
No. of terms left in Round 5) 29 - (9+1) = 19 ................since 9*3 = 27
No. of terms left in Round 6) 19 - (6+1) = 12
No. of terms left in Round 7) 12 - (3+1) = 8
No. of terms left in Round 8) 8 - ( 2+1) = 5
No. of terms left in Round 9) 5 - (1+1) = 3
**********************************************************************
The Final Recipe :

Now since we know the number of terms eliminated in each round as well the no. of terms remaining in subsequent round we can start with the process.

Let us write down the number tags of each contestant.

In Round 0), we have all 150 terms/contestants

*1 2 3 *4........................140 141 *142 143 144 *145 146 147 *148 149 150

The contestants who are to be eliminated are awarded a * , a consolation prize.

After the terms with * are struck off we proceed to

Round 1)
The terms left in this round are 100, writing them below :

*2 3 5 ............................140 *141 143 144 *146 147 149(99th term) *150(100th term)

Eliminating the terms with *, we proceed to

Round 2)
The terms left in this round should be 66. The last term which is 66th term cannot be eliminated since it is a mutiple of 66=3.22. So the term to be eliminated is 3.(22-1) = 63rd term

*3.........................................140 143 *144(64th term) 147(65th term) 149(66th term)


Round 3) : 44 terms remaining in this round

............................................. 140 143(42nd term) *147(43rd term) 149(44th term)

3.14+1 = 43rd term is the last term to be struck off.

Round 4) : 29 terms remaining in this round

.............................................. 140 *143(28th term) 149(29th term)

3.9 + 1 = 28th term is the last term to be struck off.

Round 5) : 19 terms remaining in this round

............................................... 140(18th term) *149(19th term)

3.6 + 1 = 19th term is the last term to be struck off.

Round 6) : 12 terms remaining in this round

................................................ 140(12th term)

Note that the number of terms is a multiple of 3, so the last term wont get eliminated.

Round 7) : 8 terms remaining in this round and 140 = last ie 8th term.

Note that 3.2+1 = 7th term is the last term to be struck off. So 8th term which is 140 still remains

Round 8) : 5 terms remaining in this round and 140 = last ie 5th term.

Note that 3.1+1 = 4th term is the last term to be struck off. So 5th term which is 140 still remains

Round 9) : 3 terms or contestants remaining in this round

let them be Miss Australia(with number tag X)..................Miss Wales(with number tag Y)............ and Miss China(with tag 140)

As per the strange but interesting rule of the contest, we eliminate 1st contestant and are left with

in Round 10)
Miss Wales(with tag no. Y)............ and Miss China(with tag 140)

Eliminatin again 1st contestant we are left with winner of the competition.

Miss China(with tag no. 140)

So the number tag of winner is 140........Answer
***************************************************************
Garnishing the dish now :

Important thing to note here is that there is no need to bother about the terms in the starting of the series.
After round (2), I stopped writing the numbers in the beginning of the series. I was only interested in the numbers in the last of series.
Since I knew this beforehand I made sure to write atleast last 11 numbers in the series.

How did I come to know this? Did 2,3 dry runs with few numbers and noticed that the winner is always the one who stands towards the end of the series.

****************************************

Is this the correct answer?
This method sure required a lot of brainstorming.
Do you know any other method to solve this Qn?
If you do pls share.

@ekanshtiwari said: Even i did the same question yesterday from Arun Sharma, the answer given is wrong. It has to be 28 no doubt about that.
123/5 +123/5^2 =20+4=24.....!!!

Vishal99
Rightmost non-zero digit of n!


R(n!) = Last Digit of [ 2a x R(a!) x R(b!) ]


where n = 5a + b




@[568667:Psychamour]

I find the solution convincing though the question mentions :

Several rounds happen before the final winner is selected.
That is something that bemuses me.

Could the Miss World be selected in just 7 rounds as per the rejections procedure given in the question?

puys...m stuck in middle of a problem simplified version of problem is 223^377 divided by 1000 rem?

@[403711:vishalgarg09]

Usualy such Qns are accompanied with two Qns - one asking the total number of rounds required to choose the winner and the number at which a person should stand so that he becomes the winner. This is why 'Several' has been used so that the Qn about the total number of rounds required can be asked, though the poster didnt do so.

Well, if the winner can be chosen in only 7 rounds then my friend it will save lot of operating costs and to the contestants' relief, will end up the killing suspense in their hearts sooner.

Pls unfold the mystery and post the procedure you have in ur mind for doing the whole activity in 7 rounds. It would be interesting to see some other method to solve this Qn.

[1/(1!*19!) + 1/(3!*17!) + 1/(5!*15!) +................ 1/(15!*5!) + 1/(17!*3!) +1/(19!*1!) = (2^a)/b!. find the summation of a+b. I just did a bit of it.. like since the denominator is in factorial form, I multiplied and divided the whole thing with 20!. so it's like 1/20![ 20!/(1!*19!) +...] which is like 20C1 +20C3 +20C5.... + 20C19. so B I think is 20, but can't find the value of a.. as in how will I express this whole thing in terms of 2^ something????? can anyone please help? THANK YOU.. 😃

@koyal1990 said:[1/(1!*19!) + 1/(3!*17!) + 1/(5!*15!) +................ 1/(15!*5!) + 1/(17!*3!) +1/(19!*1!) = (2^a)/b!. find the summation of a+b. I just did a bit of it.. like since the denominator is in factorial form, I multiplied and divided the whole thing with 20!. so it's like 1/20![ 20!/(1!*19!) +...] which is like 20C1 +20C3 +20C5.... + 20C19. so B I think is 20, but can't find the value of a.. as in how will I express this whole thing in terms of 2^ something????? can anyone please help? THANK YOU..
You almost solved the Qn. just this minor thing skipped your mind.
Recall that
20C0 + 20C1 + 20C2 + 20C3 + 20C4 + ...... + 20C19 + 20C20 = 2^20
and
20C1 + 20C3 + 20C5 + ..... 20C19 = 20C0 + 20C2 + 20C4 + ....... + 20C20 = (2^20)/2
So, like u did, dividing and multiplying by 20! the given series will condense into
[ 1/(1!*19!) + 1/(3!*17!) + 1/(5!*15!) +................ 1/(15!*5!) + 1/(17!*3!) +1/(19!*1!) ]
= [1/20!] [ 1/(1!*19!) + 1/(3!*17!) +..........+ 1/(5!*15!) ]
= [1/20!] [ (2^20)/2 ]
= [ (2^19)/20! ]
a =19, b= 20.
***************************
Edited :

After the insight provided by @ashish0102 in his post.

@Psychamour said:
You almost solved the Qn. just this minor thing skipped your mind.
Recall that
20C0 + 20C1 + 20C2 + 20C3 + 20C4 + ...... + 20C19 + 20C20 = 2^20
and
20C1 + 20C3 + 20C5 + ..... 20C19 = 20C0 + 20C2 + 20C4 + ....... + 20C20 = (2^20)/2
So, like u did, dividing and multiplying by 20! the given series will condense into [ 1/(1!*19!) + 1/(3!*17!) + 1/(5!*15!) +................ 1/(15!*5!) + 1/(17!*3!) +1/(19!*1!) ] = [1/20!] [2] [ 1/(1!*19!) + 1/(3!*17!) +..........+ 1/(5!*15!) ] ..................( since we have two identical terms for each term ) = [1/20!] [2] [ (2^20)/2 ] = [ (2^20)/20! ] a = b= 20.
pls,can you check the answer?
I am a lil confused.. 1stly whn we are taking away the 20c0, 20c2, 20c4 and all the even ones, then why are we just dividing by 2?????
@koyal1990 said:
I am a lil confused.. 1stly whn we are taking away the 20c0, 20c2, 20c4 and all the even ones, then why are we just dividing by 2?????
You have almost made it ....

As the question ask :
we can reduce the info given to following form (as you did)
Let dividing and multiplying by 20!
1/20!{ 20C1, 20C3........} = X/20!
NOW , We know : 20C0+20C1+20C2+........20C20= 2^20
SO, lets divide it in two half :
{20C0+ 20C2+20C4+......+20C20} +{20C1+20C3+20C5+....20C19} = 2^20
2^19 + X = 2^20
or, X = 2^20-2^19
So, X/20! = 2^19/20! i.e a= 19 and b=20...


3392 RepliesGo to:
◄ 1 2 168 169 170
just a lil more help.. I am just not understanding the 2^19 part!! It's just not registering.. I know it is stupid but cud you please explain it in a bit more detailed fashion? I know that 20c0+ 20c1+.... 20c20= 2^20. but why is 20c1+20c3+20c5+... 20c19 =2^19??? please if you could elaborate on this a bit..!!
@ashish0102 said:
You have almost made it ....
As the question ask :
we can reduce the info given to following form (as you did)
Let dividing and multiplying by 20!
1/20!{ 20C1, 20C3........} = X/20!
NOW , We know : 20C0+20C1+20C2+........20C20= 2^20
SO, lets divide it in two half :
{20C0+ 20C2+20C4+......+20C20} +{20C1+20C3+20C5+....20C19} = 2^20
2^19 + X = 2^20
or, X = 2^20-2^19
So, X/20! = 2^19/20! i.e a= 19 and b=20...
B
I
U
img
:)
    ashish0102
    3392 RepliesGo to:
    ◄ 1 2 168 169 170
    just a lil more help.. I am just not understanding the 2^19 part!! It's just not registering.. I know it is stupid but cud you please explain it in a bit more detailed fashion? I know that 20c0+ 20c1+.... 20c20= 2^20. but why is 20c1+20c3+20c5+... 20c19 =2^19??? please if you could elaborate on this a bit..!!
    @ashish0102 said:
    You have almost made it ....
    As the question ask :
    we can reduce the info given to following form (as you did)
    Let dividing and multiplying by 20!
    1/20!{ 20C1, 20C3........} = X/20!
    NOW , We know : 20C0+20C1+20C2+........20C20= 2^20
    SO, lets divide it in two half :
    {20C0+ 20C2+20C4+......+20C20} +{20C1+20C3+20C5+....20C19} = 2^20
    2^19 + X = 2^20
    or, X = 2^20-2^19
    So, X/20! = 2^19/20! i.e a= 19 and b=20...
    B
    I
    U
    img
    :)
      ashish0102

      just a lil more help.. I am just not understanding the 2^19 part!! It's just not registering.. I know it is stupid but cud you please explain it in a bit more detailed fashion? I know that 20c0+ 20c1+.... 20c20= 2^20. but why is 20c1+20c3+20c5+... 20c19 =2^19??? please if you could elaborate on this a bit..!!

      @ashish0102 said:
      You have almost made it ....
      As the question ask :
      we can reduce the info given to following form (as you did)
      Let dividing and multiplying by 20!
      1/20!{ 20C1, 20C3........} = X/20!
      NOW , We know : 20C0+20C1+20C2+........20C20= 2^20
      SO, lets divide it in two half :
      {20C0+ 20C2+20C4+......+20C20} +{20C1+20C3+20C5+....20C19} = 2^20
      2^19 + X = 2^20
      or, X = 2^20-2^19
      So, X/20! = 2^19/20! i.e a= 19 and b=20...
      @ashish0102 said:
      You have almost made it ....
      As the question ask :
      we can reduce the info given to following form (as you did)
      Let dividing and multiplying by 20!
      1/20!{ 20C1, 20C3........} = X/20!
      NOW , We know : 20C0+20C1+20C2+........20C20= 2^20
      SO, lets divide it in two half :
      {20C0+ 20C2+20C4+......+20C20} +{20C1+20C3+20C5+....20C19} = 2^20
      2^19 + X = 2^20
      or, X = 2^20-2^19
      So, X/20! = 2^19/20! i.e a= 19 and b=20...



      Using Binomial : nc0 +nc2 +....=2^n-1
      @Mad-Head said: Questionnaire: If then the remainder when is divided by 7 isOptions:a) 3 b) 0 c) None d) 4 If then is always divisible byOptions:a) 13 b) None c) 15 d) 14 If then is always divisible byOptions:a) None b) 13 c) 15 d) 17 If then the remainder when is divided by 7 isOptions:a) 5 b) 0 c) None d) 4 Which of the following is a multiple of5, ?Options:a) n^6-n^2+10 b) n^{15}-n^{13}+25 c) None d) n^5-n-25
      question 1 answer is 0. question 2 answer is 13. 3 and 4 is kind of illegible. 5 not sure but I think it is option D. not too sure though!!

      just a lil more help.. I am just not understanding the 2^19 part!! It's just not registering.. I know it is stupid but cud you please explain it in a bit more detailed fashion? I know that 20c0+ 20c1+.... 20c20= 2^20. but why is 20c1+20c3+20c5+... 20c19 =2^19??? please if you could elaborate on this a bit..!!

      @ashish0102 said:
      You have almost made it ....
      As the question ask :
      we can reduce the info given to following form (as you did)
      Let dividing and multiplying by 20!
      1/20!{ 20C1, 20C3........} = X/20!
      NOW , We know : 20C0+20C1+20C2+........20C20= 2^20
      SO, lets divide it in two half :
      {20C0+ 20C2+20C4+......+20C20} +{20C1+20C3+20C5+....20C19} = 2^20
      2^19 + X = 2^20
      or, X = 2^20-2^19
      So, X/20! = 2^19/20! i.e a= 19 and b=20...
      @ashish0102 said: just a lil more help.. I am just not understanding the 2^19 part!! It's just not registering.. I know it is stupid but cud you please explain it in a bit more detailed fashion? I know that 20c0+ 20c1+.... 20c20= 2^20. but why is 20c1+20c3+20c5+... 20c19 =2^19??? please if you could elaborate on this a bit..!!
      Using Binomial : nc0 +nc2 +....=2^n-1
      oh yaaaaaaa.... oh God!! just so stupid of me.. I guess it happens when to try to stuff too much inside your brain, the size of which is almost microscopic.. thanks for the help mate!! Can't believe I did the whole thing almost n got stuck here.. 😞

      just a lil more help.. I am just not understanding the 2^19 part!! It's just not registering.. I know it is stupid but cud you please explain it in a bit more detailed fashion? I know that 20c0+ 20c1+.... 20c20= 2^20. but why is 20c1+20c3+20c5+... 20c19 =2^19??? please if you could elaborate on this a bit..!!

      @ashish0102 said:
      You have almost made it ....
      As the question ask :
      we can reduce the info given to following form (as you did)
      Let dividing and multiplying by 20!
      1/20!{ 20C1, 20C3........} = X/20!
      NOW , We know : 20C0+20C1+20C2+........20C20= 2^20
      SO, lets divide it in two half :
      {20C0+ 20C2+20C4+......+20C20} +{20C1+20C3+20C5+....20C19} = 2^20
      2^19 + X = 2^20
      or, X = 2^20-2^19
      So, X/20! = 2^19/20! i.e a= 19 and b=20...
      @koyal1990 said: just a lil more help.. I am just not understanding the 2^19 part!! It's just not registering.. I know it is stupid but cud you please explain it in a bit more detailed fashion? I know that 20c0+ 20c1+.... 20c20= 2^20. but why is 20c1+20c3+20c5+... 20c19 =2^19??? please if you could elaborate on this a bit..!!
      oh yaaaaaaa.... oh God!! just so stupid of me.. I guess it happens when to try to stuff too much inside your brain, the size of which is almost microscopic.. thanks for the help mate!! Can't believe I did the whole thing almost n got stuck here..



      It's Happens sometimes ...
      All d best

      what is the remainder when 30^20^10^5 is divided by 19???