PG Premier League - Chennai Dream Team

[krsh.vik]

Originally Posted by satanica

You have an infinte supply of notes of Rs.6, Rs.9 and Rs.20.
Now a guy will come and ask you for some money. Obviously you can not give him every amount that he asks for with these set of notes. e.g. You can not give him Rs.7, Rs.8, Rs.11 etc.

What is the maximum amount that cannot be given with these notes?

43 Shall post it if someone else doesn't

[shivam_01]

Originally Posted by krsh.vik

43 Shall post it if someone else doesn't

There can be many others like u take the example of 51 ??
I guess there might be some condition invloved in this case as the infinte supply of notes of Rs.6, Rs.9 and Rs.20 can give n number of options .

I think there must be some excahnge involved happening
@satanica answer post kar do bhai

[krsh.vik]

Originally Posted by shivam_01

There can be many others like u take the example of 51 ??
I guess there might be some condition invloved in this case as the infinte supply of notes of Rs.6, Rs.9 and Rs.20 can give n number of options .

I think there must be some excahnge involved happening
@satanica answer post kar do bhai

51 is possible buddy

9*5 + 6*1

[monsterkartik]

Originally Posted by naga25french

let me make a wild guess on this question..

the maximum amount amount that cannot be given is

(LCM of 6,9,20) - 1

180 - 1 = 179...

so answer should be 179..

looks weird...

but i guess this should be maximum amount...

this is toooo wild a guess

BTW 179 is payable as

20*4 = 80
9*11 = 99

Total = 179



Confused

Cheers

MK

[monsterkartik]

Originally Posted by krsh.vik

51 is possible buddy

9*6 + 6*1

correction

9*5 + 6*1



Cheers

MK

[nishant_rungta]

Originally Posted by an_elusive_lady

what is this passage on women n pillows n tears?? chauvinistic! i protest.

Ok on a very lighter note read these

[krsh.vik]

Originally Posted by monsterkartik

this is toooo wild a guess

BTW 179 is payable as

20*4 = 80
9*11 = 99

Total = 179

I think there can be many solutions for this, eg 43, 61, 79 etc

Confused

Cheers

MK

Dude 61 = 20*2 + 9 + 2*6
79 = 20*2 + 9 + 6*5

Solution :

6 mod 6 = 0 which means all multiples of six are possible
9 mod 6 = 3 which means all numbers of the form 6x+3 >= 9 are possible
20 mod 6 = 2 6x+2 >= 20 are possible
29 mod 6 = 5 6x+5 >= 29 are possible
40 mod 6 = 4 6x+4 >= 40 are possible
49 mod 6 = 1 6x+1 >= 49 are possible

answer is 43

[satanica]

Originally Posted by krsh.vik

Dude 61 = 20*2 + 9 + 2*6
79 = 20*2 + 9 + 6*5

Solution :

6 mod 6 = 0 which means all multiples of six are possible
9 mod 6 = 3 which means all numbers of the form 6x+3 >= 9 are possible
20 mod 6 = 2 6x+2 >= 20 are possible
29 mod 6 = 5 6x+5 >= 29 are possible
40 mod 6 = 4 6x+4 >= 40 are possible
49 mod 6 = 1 6x+1 >= 49 are possible

answer is 43

That's right ...

This following could have been a better way.

Each number can be represented as 3x, 3x+1 or 3x+2.
9-6 mod 3 = 0 and 6 mod 3 = 0 means all 'n' of form 3x are possible for 3x>=6

Since 6 and 9 are both multiples of 3, 'n' of form 3x+1 and 3x+2 will have atleast a 20 in them.
20 mod 3 = 2 means all 'n' of form 3x+2 are possible for 3x+2>=26
20*2 mod 3 = 1 means all 'n' of form 3x+1 are possible for 3x+1>=46

The maximum possible value of number not possible is of form 3x+1 and is less than 46 i.e. 43.

Another problem.

How many ways are there to express 1000000 as a product of exactly three integers greater than 1? (For the purpose of this problem, abc is not considered different from bac, etc.)

[krsh.vik]

Originally Posted by satanica

That's right ...

Another problem.

How many ways are there to express 1000000 as a product of exactly three integers greater than 1? (For the purpose of this problem, abc is not considered different from bac, etc.)

Not an easy one ! Unable to figure out a method to avoid repetitions

Nice questions satan Keep 'em coming

[naga25french]

Originally Posted by krsh.vik

Not an easy one ! Unable to figure out a method to avoid repetitions

Nice questions satan Keep 'em coming

superb question ..........

10^6 = 10 * 10^2 * 10 ^3

10 = 2*5

100 = 2*2*5*5

1000 = 2*2*2*5*5*5


the number of triplet product is ( 2 * 4 * 6 * 3) = 144

correct me if i am wrong..