Riddles Options

[shivani]

no frigging idea.. Give some clue man!
I see our mathemagician around.. maybe he would have some answer

[vikram]

Ok here is the clue....!!
" you have to think the way computer think's "

[vivekpm]

Ok here is the clue....!!
" you have to think the way computer think's "

Answer this. You mean 1's and 0's?

Cheers,

[vikram]

Ok here is the clue....!!
" you have to think the way computer think's "

Answer this. You mean 1's and 0's?

Cheers,

oh man..i think you will get this also ......

[vivekpm]

19? Write the number in binary and add number of 1's to total number of digits in binary representation of the number?

13 = 1101 = 4 digits + 3(number of 1's) = 7.
9304 = 10010001011000 = 14 digits + (5 1's) = 19

Cheers,

[vikram]

19? Write the number in binary and add number of 1's to total number of digits in binary representation of the number?

13 = 1101 = 4 digits + 3(number of 1's) = 7.
9304 = 10010001011000 = 14 digits + (5 1's) = 19

Cheers,

Great Vivek..i thought its a tuff one ...but you did it again....

Great !!

[shivani]

ok here is another one.. the kind Vivek likes and Vikram loves : )
(and rest of us hate..not sure about hits though)
Make 29 out of 3, 4, 5, and 8.

[vivekpm]

19? Write the number in binary and add number of 1's to total number of digits in binary representation of the number?

13 = 1101 = 4 digits + 3(number of 1's) = 7.
9304 = 10010001011000 = 14 digits + (5 1's) = 19

Cheers,

Great Vivek..i thought its a tuff one ...but you did it again....

Great !!

Your clue helped. I would have taken time to move to other bases. But since you told binary, I immediately jumped to it.

Cheers,

[vivekpm]

ok here is another one.. the kind Vivek likes and Vikram loves : )
(and rest of us hate..not sure about hits though)
Make 29 out of 3, 4, 5, and 8.

Looks like this one is easier (If I am not missing anything that is)

5 * (8-3) + 4 = 5 * 5 + 4 = 25 + 4 = 29?

Cheers,

[vivekpm]

Continuing Vikram's coin theme, here is one more...

You have 5 boxes full of identical coins. One of the box has all defective coins. One non-defective coin weigh 10 gm while defective one weigh 9 gm. How will find the box with defective coins using a weighing pan only once?

Cheers,

[amicable]

Any pointers/hints?

[vivekpm]

Any pointers/hints?

Sorry somehow missed your post.

Hint: You have to weigh samples from different boxes together

Cheers,

[amicable]

Ok.. Here's My Guess:

I will label boxes as A, B, C, D and E
and each coin in each box as 1A, 2A, 3A... for set A, 1B, 2B, 3B for set B, etcc.. for coins in all the 5 boxes.

Now, for simplicity of calculation (and also because I like the tables of 10 and 9 ) I will take:

1 coin from box A
2 coins from box B
3 coins from box C
4 coins from box D
and no coins from box E


And I will keep these coins on weighing pan. From the readings, I will identify which box has defective coins. Here's how:

Box A is defective if reading is 99 because:
9A + (2 * 10)B + (3 * 10)D + (4 * 10)D
9 + 20 + 30 + 40
= 99


Similarly,
Box B is defective if reading is 98 because:
10 + 18 + 30 + 40 = 98


Box C is defective if reading is 97 because:
10 + 20 + 27 + 40 = 97


Box D is defective if reading is 96 because:
10 + 20 + 30 + 36 = 96


Box E is defective if reading is 100 because:
10 + 20 + 30 + 40 = 100



( Assumption: Each box has atleast 4 coins )


Is this approach correct?

[Mandrake]

There!! One more friend lost to maths
Amita, when will I come near you all?

Mere toh khopdi hi nahi chalti

[amicable]

Arre Mandrake, yeh to Vivek ka hint helped me out, otherwise, I was lost in numbers.

I am not a mathemagician like Vivek and others here.. And since no one replied till date, so that gave me ample time to solve it, warna to solve hua nahi hua aur solutions bhi aa jaate hai...