hgupta

[hide]reverse alphabetical?[/hide]

It seems so.

I worked this out last night (one of those "I woke up with the answer deals).

 

If anyone wants the answer just PM me so others can keep trying

Any hints??

several pages back, there's a hint that helped me figure it out.

No worries, finished all!!

Cheers!!

Already cleared that level!! Just asking if there is any relation between the answer and "His Project".

BTW, now stuck at level 8. Tried nearly 25 names of Saints!!

Does the answer of level 7 make any sense??

sure... answer it correctly.

Thank you

BTW, still

Once you get to here:

 

 

 

you can figure out what A' date=' B, C, X, Y and Z are... right?[/quote']

thank you ecoli!! that was really simple.

 

Any suggestions for Level 7?

Do you know the answer? Is it 18?

even with the hint, I am still stuck at Page 2

hehe, thank you!!

actually did 3/7 for the second alphabet (or first vowel) !!

hello, i could never work out this problem, and i am fed up it now,

Given a word "ARTICLE", its alphabets are jumbled up and then placed at random. What is the probability that the vowels will end up taking even places?

hope my question is clear!!

thank you

thank you!!

Hello,

I wanted to try the LateX codes, that are used by members here to write Maths formulae etc. Where could I try them?? Is there any Test Area forum available here?

Thank you

Since CAB=30 we can say DAB=60

it should be 105^o.

taking tri(ABC), given ang(BAC) = 30 and AB = BC, so ang(ACB) = ang(ABC).

sum of all angles of a triangle = 180^o

therefore, ang(ABC+ACB+BAC) = 180^o and BAC = 30^o

=> 2*ang(ACB) = 150^o since ACB = ABC

=> ACB = 75^o

=> ang(CAD) = ang(ACB) = 75^o, property of ||ogram, alternate angles of a diagnol are equal

=> BAD = BAC + CAD = 30^o + 75^o = 105^o

also, BAD = ADE, property of || lines

therefore, ADE = 105^o

WAIT, could you please re-read and re-post your question!! if AC = AB then it is no longer a ||gram but a quadrilateral!!

Besides what DAK and Drug Addict( ) posted, AB when prescribed for viral infections weakens your defense mechanism, thus making you prone to infections, and increases the rate of risk of the infection.

plus, an adult (assuming healthy) may not expose to side effects, but AB when given to children, old-aged people, pregnant women, increases that risk also. in most of the cases, these people do show some kind of abnormal behavior due to some dosage of ABs.

HI..I need some reference sites towards some of the basic coding algorithms' date=' viz. the Huffman and arithmetic coding techniques.

 

 

Thanks in advance[/quote']

Huffman Coding - PlanetMath.org

 

Huffman Coding - mathworld.wolfram.com

And does reading the book help in any way to solve the encryption?

 

I haven't read it - so do I still have an equal change as someone else who has read the book at solving this?

i just wanted to know if the question does not fall into a category described (or used) in the book.

and it's confirmed now.

yes Luke is correct. you need some more information in order to solve it. i tried using the Trig., and Similar triangles concept but to no avail.

please see if you are not missing any point.

i haven't read Digital Fortress till now, so just asking a small question.

is the answer a word, or a number or some random text or something else?

i dint understand.pls.explain

umm, i think that's the whole explanation.

well if you could not understand a/some specific step(s), post back and i will try to redo it again in a different manner.

got the solution, posting here if someone else might want to know:

Since A to I are 1 to 9, then: A+B+C+D+E+F+G+H+I = 45 (eq_1)

 

We know that: A+B+C+D + D+E+F+G + G+H+I = 51 (17 * 3) (eq_2)

 

Therefore the difference between the eq_1 and eq_2 is D+G, so D+G = 51-45 = 6

 

3 pairs of numbers can add together to form 6: 1 & 5, or 2 & 4, or 3 & 3

 

The correct pair must be 1 and 5 because we know A = 4, and repetition of 3.

 

If we guess that D=1 and G=5, then 4+B+C+1 = 17 and 5+H+I = 17

 

This means that B+C = 17-4-1 = 12, and H+I = 17-5 = 12.

 

Three pairs of numbers add together to make 12: 3+9, 4+8, and 5+7.

 

Of these, it can't be 4+8 because A=4, and it can't be 5+7 because we know that either D or G = 5, so the correct pair must be 9+3.

 

But, we have two pairs of numbers that need to add up to 12: B+C and H+I, and only one pair of numbers to acheive it with, therefore our original assumption that D=1 and G=5 must be wrong.

 

Now assume that D = 5 and G = 1:

 

We now have 4+B+C+5 = 17, and 1+H+I = 17.

 

So, B+C = 17-4-5 = 8, and H+I = 17-1 = 16.

 

Only one pair of numbers add up to make 16: 7+9, so H+ I = 7+9

 

Of the remaining numbers one pair adds up to 8: 2+6. Therefore B+C = 2+6

 

We are only left with one other pair: 3+8, so E+F must = 3+8.

 

If you now plug all these numebers back into the original equations:

 

4 + 2 + 6 + 5 = 17

 

5 + 3 + 8 + 1 = 17

 

1 + 7 + 9 = 17.

 

 

Therefore D = 5, and G =1

Yes the answer is 32. forgot to add the 12 wins of the friend.

sorry about my previous answer. is it 48??

my bad, thought that a dice has 8 faces!!

6 faces, so 6 options for 1 and 6 on opposite sides, consecutively 4 options for 2 and 5, and so 2 options for 3 and 4.

am i right??

My friend came to my home for vacation. Every day we played a chess game.Where the looser should give the winner a choclate.Finally one day my friend gave me 8 choclates even though she won 12 games.How many days did she stay in my home?

sorry for such posts!! i am preparing for few exams and these questions are killing me because i make very silly mistakes in Maths.

kindly bear with me!!

There is a six faced die.Numbering should be done from 1 to 6.How many ways the numbering can be done , with a condition that 1 and 6, 2 and 5 , 3 and 4 comes on the opposite sides of the die.