[saloni30]

no its not 11. the ans is 29.

[chetna]

Originally Posted by saloni30

no its not 11. the ans is 29.

yes,have edited my answer...though its still not 29 :(

[warrior]

Originally Posted by saloni30

A set S consists of 143 natural nos, each of which is a perfect cube. the maximum no of elements of S that one can always find such that each of them leaves the same remainder when divided by 13 is

1)11 2)24 3)29 4)36

see

if we take cube of first 13 natural nos we get

rem 0 - 13
1- 1,3,9
5- 7,8
8 - 2,5,6,11
12- 4,10,12

now we have 5 different groups. and any no can we written as 13k + p ( where p is 1 to 12)

now if we distrubute our nos such that no of natural nos with the same reminder are same we have 28 nos in each. I mean we have 28 nos with 0 rem 1 no as 0 and so on.
but three group will have 29 nos in them so 29 is the ans

simply put it is max integer greater than (143/5) =29

that's my 800

[chetna]

Originally Posted by warrior

see


8 - 2,5,6,11

thanks ...........

i missed out 11^3%13=8

[fountainhead82]

Originally Posted by warrior

see

if we take cube of first 13 natural nos we get

rem 0 - 13
1- 1,3,9
5- 7,8
8 - 2,5,6,11
12- 4,10,12

now we have 5 different groups. and any no can we written as 13k + p ( where p is 1 to 12)

now if we distrubute our nos such that no of natural nos with the same reminder are same we have 28 nos in each. I mean we have 28 nos with 0 rem 1 no as 0 and so on.
but three group will have 29 nos in them so 29 is the ans

simply put it is max integer greater than (143/5) =29

that's my 800

--------------------------------------------------
Okay Maybe I am too stupid but I don't really get this question or the solution to it.


@warrior -- First of all, there is a mistake here
11^3 (mod 13) = 5 (not 8 )
so you have three sets
rem 0 - 13
1- 1,3,9
5- 7,8,11
8 - 2,5,6
12- 4,10,12

which means, IF you take only cubes of consecutive numbers, you can have 3 X 11 = 33 (Maximum numbers)
For example for remainder 1
you can have numbers of the form (13k+1), (13k+3) and (13k+9) = {1,3,9,14,17,19,...139}

However there is no such restriction outlined by the question(for taking cubes of consecutive numbers).
Therefore you CAN pick numbers at your will
if you pick numbers of the form (13k+1)^3 only, then you can have all 143 numbers resulting in same remainder.

Is there something missing or implicit in the question which is too obvious to all but me?

Let me know, I think I my brain is scrambled from all the math I have been doing lately.

I tried this approach too:
Picking 143 numbers = picking 143 perfect cube roots

if two numbers have same remainder, then there diff should be div by 13.
It means (a^3 - b^3) (mod 13) = 0
(a-b)(a^2+b^2+ab) is div by 13

one of them should be div by 13 -->
(a-b) = 13k
or (a^2+b^2+ab) = 13k

I couldn't go anywhere from here,
Help anyone?

[nagi_nov3]

Originally Posted by warrior

see

if we take cube of first 13 natural nos we get

rem 0 - 13
1- 1,3,9
5- 7,8
8 - 2,5,6,11
12- 4,10,12

now we have 5 different groups. and any no can we written as 13k + p ( where p is 1 to 12)

now if we distrubute our nos such that no of natural nos with the same reminder are same we have 28 nos in each. I mean we have 28 nos with 0 rem 1 no as 0 and so on.
but three group will have 29 nos in them so 29 is the ans

simply put it is max integer greater than (143/5) =29

A set S consists of 143 natural nos, each of which is a perfect cube. the maximum no of elements of S that one can always find such that each of them leaves the same remainder when divided by 13 is

1)11 2)24 3)29 4)36



Hi,
I'm still not convinced with ur explanation :(

if we take non consecutive numbers
all as 13 multiples like 13^3,26^3...........

we will have 143 numbers that will give same remainder..

if that is not the case
1^3,2^3 .......till 13^3 mod 13 will gives

1(1),8(2),1(3),12(4),8(5),8(6),5(7),5(8 ),1(9),12(10),5(11),12(12),0(13)

0 - 13
1- 1,3,9
5- 7,8,11 (11^3 mod 13 = -2^3 mod 13 = -8 mod 13 = 5 mod 13)
8- 2,5,6,
12 - 4,10,12

so we can have max 3(any of the 1,5,8,12) for 1 to 13

we need max so we can have 11*3 = 33

u can't generalize as 143/5 (becuase if all remainders are equal in number then only we can divide)

so answer must be 33.

[arun_bisht]

Originally Posted by vineet.nitd

No , here u have to find the remainder when the number is divided by 9 .

that wud come to be 7 which is the digit sum ....

Hi vineet,

i too, got th eans. as 7,

[warrior]

Originally Posted by nagi_nov3

A set S consists of 143 natural nos, each of which is a perfect cube. the maximum no of elements of S that one can always find such that each of them leaves the same remainder when divided by 13 is

1)11 2)24 3)29 4)36



Hi,
I'm still not convinced with ur explanation :(

if we take non consecutive numbers
all as 13 multiples like 13^3,26^3...........

we will have 143 numbers that will give same remainder..

if that is not the case
1^3,2^3 .......till 13^3 mod 13 will gives

1(1),8(2),1(3),12(4),8(5),8(6),5(7),5(8 ),1(9),12(10),5(11),12(12),0(13)

0 - 13
1- 1,3,9
5- 7,8,11 (11^3 mod 13 = -2^3 mod 13 = -8 mod 13 = 5 mod 13)
8- 2,5,6,
12 - 4,10,12

so we can have max 3(any of the 1,5,8,12) for 1 to 13

we need max so we can have 11*3 = 33

u can't generalize as 143/5 (becuase if all remainders are equal in number then only we can divide)

so answer must be 33.

see you have 5 different groups for reminder

0 - 13
1- 1,3,9
5- 7,8,11
8- 2,5,6,
12 - 4,10,12

now if I need to choose 10 natural nos I can always choose 2 nos which give me 0 as rem, 2 nos 1 as rem, and so on

for 143 I can chosse 28 nos of each type ( the type of rem they leave) now among the three left I have to choose three nos belonging to 3 different groups.

[nagi_nov3]

Originally Posted by warrior

see you have 5 different groups for reminder

0 - 13
1- 1,3,9
5- 7,8,11
8- 2,5,6,
12 - 4,10,12

now if I need to choose 10 natural nos I can always choose 2 nos which give me 0 as rem, 2 nos 1 as rem, and so on

for 143 I can chosse 28 nos of each type ( the type of rem they leave) now among the three left I have to choose three nos belonging to 3 different groups.

Here the thing is we need to select the set such that it gives max. type of numbers which give same remainder.

for 10 numbers
also we can have max 3. ( 1,3,9)

for 143 we can select 28 as well but we need max numbers which will us same remainder so that will become 33.

[warrior]

Originally Posted by nagi_nov3

Here the thing is we need to select the set such that it gives max. type of numbers which give same remainder.

for 10 numbers
also we can have max 3. ( 1,3,9)

for 143 we can select 28 as well but we need max numbers which will us same remainder so that will become 33.

qn states

A set S consists of 143 natural nos, each of which is a perfect cube. the maximum no of elements of S that one can always find such that each of them leaves the same remainder when divided by 13 is

we are more than sure of 29 are u sure that 33 nos will always be there