[varun nakra1]

post deleted

[dreams_hemant]

hi aarav great thread


well i sincerely feel that a thread of such a nature can help us a lot in thinking abt a prob in different ways................

welll i think a thread on VA can also help us a lot........

anywayz i would like to post a beautiful prob(my perception)

u went to a show called KHUL JA SIM SIM and the host(HUSSAIN) asked u to select one of the 3(closed) doors. After u have selected a door... he opened a door in which there was no prize .... (mind u there is a prize in only 1 door). He always opens a door in which there is no prize

now he gives u an option to change the door u selected.....

wwhat would u do?? will u switch to the other door or stick with ur own choice.... which one can be more profitable

[amar_kashyap]

Originally Posted by dreams_hemant

hi aarav great thread


well i sincerely feel that a thread of such a nature can help us a lot in thinking abt a prob in different ways................

welll i think a thread on VA can also help us a lot........

anywayz i would like to post a beautiful prob(my perception)

u went to a show called KHUL JA SIM SIM and the host(HUSSAIN) asked u to select one of the 3(closed) doors. After u have selected a door... he opened a door in which there was no prize .... (mind u there is a prize in only 1 door). He always opens a door in which there is no prize

now he gives u an option to change the door u selected.....

wwhat would u do?? will u switch to the other door or stick with ur own choice.... which one can be more profitable

Hemant,
Plz read the instructions at the start of the thread. You gotta solve a problem ...at least give a method....then only you post the new problem. And yeah, you have to post the problem specifying its number. I think u have not gone thru the previous pages. Moreover the above question is more of a DI and LR type, i think there is already a thread on puzzles. This problem will be better there...

Hope you don't mind .... All this is required to keep track of problems and maintain order.

Note: plz edit your post.....and try solving the problem no 17

[amar_kashyap]

Since nobody is posting a new problem (of all those who tried solving it..)..I will do the service..
Here we go

Problem 18
a, b, and c are three natural numbers such that a+b+c is divisible by 3. It can be certainly concluded about (a^3+b^3+c^3) that it is divisible by:

A. 5 B. 9 C.3 D. None of the previous three


Note: It's not tough..

[vineet.nitd]

Originally Posted by amar_kashyap

Since nobody is posting a new problem (of all those who tried solving it..)..I will do the service..
Here we go

Problem 18
a, b, and c are three natural numbers such that a+b+c is divisible by 3. It can be certainly concluded about (a^3+b^3+c^3) that it is divisible by:

A. 5 B. 9 C.3 D. None of the previous three


Note: It's not tough..

we know that ,

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - ac - bc)

taking a+b+c=3k , where k is an integer

=> a3 + b3 + c3 = 3k*(a2 + b2 + c2 - ab - ac - bc) + 3abc

=>a3 + b3 + c3 = 3[k(a2 + b2 + c2 - ab - ac - bc)+abc]

so it is a multiple of 3....

and hence option (c) is correct .............

[amar_kashyap]

Originally Posted by vineet.nitd

we know that ,

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - ac - bc)

taking a+b+c=3k , where k is an integer

=> a3 + b3 + c3 = 3k*(a2 + b2 + c2 - ab - ac - bc) + 3abc

=>a3 + b3 + c3 = 3[k(a2 + b2 + c2 - ab - ac - bc)+abc]

so it is a multiple of 3....

and hence option (c) is correct .............

U r right boss!!..
Now tell...what if we have 'n' natural numbers a1, a2, a3, a4...............an such that
a1+a2+a3+a4+........an is divisible by '3'. What can u conclude about (a1)^3+(a2)^3+(a3)^3+...........................+( an)^3...

By the way u post a new problem with no.....

[vineet.nitd]

Originally Posted by amar_kashyap

U r right boss!!..
Now tell...what if we have 'n' natural numbers a1, a2, a3, a4...............an such that
a1+a2+a3+a4+........an is divisible by '3'. What can u conclude about (a1)^3+(a2)^3+(a3)^3+...........................+( an)^3...

By the way u post a new problem with no.....

that will also be a multiple of 3 ...always......

[amar_kashyap]

Originally Posted by vineet.nitd

that will also be a multiple of 3 ...always......

Can u show that?..

Now both of us are having equal no of posts..

Why dont u post a new problem..plzzz!!!

[charly]

Originally Posted by chautauqua

let the 11th, 12th, 13th and 14th term of this AP be: a-3d, a-d, a+d, a+3d
with 2d as the common difference

=> (a^2-d^2)(a^2-9d^2) = k^2 - 256=(k+16)(k-16)
=> 8d^2 = 32
=> d=2 => 2d = 4 .... answer is option c)

While offering my sincerest apologies for breaking the thread flow, i request someone to please clarify how the step 8d^2=32 was arrived at in above soln.
thank u.

[vineet.nitd]

Problem 19 #

There are 6 boxes numbered 1, 2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is ......
(1) 5 (2) 21 (3) 33 (4) 60