If U can.....answer this

[rkkarnani]

It can be 7,1,1 i.e. 7X1X1=7

 

[anuragmunjal]

If so the lucky number would be 7X7X7

yep 7*7*7= 343 should be the lucky number

 

[anuragmunjal]

yep 7*7*7= 343 should be the lucky number

if each of the other two cards has a value lower than 7, then obviously this is wrong...

beaten....

 

[rkkarnani]

Question: Lucky fall

George and Harry enjoy creating and solving puzzles. They also enjoy playing card games. On one occasion, George was shuffling the cards, a standard deck, when three cards fell out and landed face-down on the table. George picked up the cards and was about to return them to the deck when he noticed an interesting fact about them.

He said, "I have three cards here, Harry. As luck would have it, if I take the numerical value of each card and multiply the three values together, the result equals my lucky number. The highest card is a seven. What are the other two cards?"

Harry considered the problem for a few seconds and replied, "I suppose that an ace has a value of one, so each of your other two cards has a value lower than seven and possibly as low as one, right?"
"Right," confirmed George.
"Well," added Harry, "you haven't told me what your lucky number is, so I could only guess what the two cards are. I need more information to work it out."
"I see your difficulty," replied George, "so I will give you one more clue: the lowest card is a spade."
Harry pondered the problem again for a while, and then correctly identified the two cards.

What were they?

Seems there can be N combinations unless there is some "catch" in the language!!!

 

[whyLately]

1. Out of all the possibilities, there are only 2 outcomes with the same result on multiplying the 3 numbers i.e (7,4,1) and (7,2,2) given that the other two numbers (excluding 7) cannot be greater than 7.

How about 7,6,1 and 7,3,2? Or 7,6,2 and 7,4,3? Providing that all 3 cards are different in denomination does not resolve these cases.

 

[whyLately]

Read the next line in my reply.

"Also, one of the 2 possibilities must have the other 2 numbers as identical"

Read now and thanks.

 

[vssoma]

1. Out of all the possibilities, we are looking for 2 outcomes with the same result on multiplying the 3 numbers given that the other two numbers (excluding 7) cannot be greater than 7. Also, one of the 2 possibilities must have the other two numbers as identical so that statement 2 below can break the tie. The possibilities are (7,4,1) and (7,2,2).

2. Assuming that the statement "the lowest card is a spade" implies that there is a unique lowest card, rules out (7,2,2)

3. The numerical value of the cards are (7,4,1) and the lucky number is 28?

I have added a question mark above because I have a problem with the puzzle statement. Without going into too much detail, I do not think the puzzle is framed properly.

:thumb: :clap:

 

[vssoma]

Question: naughty word

A woman walks into a darkened room to find a derogatory name for a woman in large letters on the wall, which upset her.

Why did it bother her so?

you can form a question.....that I'll give you answer in the format of 'yes' or 'no' only....if U need clue...

 

[MurAtt]

How about 7,6,1 and 7,3,2? Or 7,6,2 and 7,4,3? Providing that all 3 cards are different in denomination does not resolve these cases.

Read the next line in my reply.

"Also, one of the 2 possibilities must have the other 2 numbers as identical"

How did u get down to selecting the 2 possibilities .. what is the logic behind these 2 numbers? Why not as whyLately has put 7, 6, 1 or 7,6,5 or any other combination? Could be anything ...

Also there is nothing special abt 28 to be a lucky number .. it could be 7,3,1=21 or for that matter any combination ...

Please do elaborate ....

 

[rkkarnani]

Question: naughty word
A woman walks into a darkened room to find a derogatory name for a woman in large letters on the wall, which upset her.

Why did it bother her so?

you can form a question.....that I'll give you answer in the format of 'yes' or 'no' only....if U need clue...

Only Owls can see in "dark"!!!