What Could Be The Product?

Zach chooses five numbers from the set {1, 2, 3, 4, 5, 6, 7} and tells their product to Claudia. She finds that this is not enough information to tell whether the sum of Zach’s numbers is even or odd. What is the product that Zach tells Claudia?

Here is the way to get that product!

Guessing The Correct Product in Question!

What was the question?

When Zach tells the product of 5 numbers that he has chosen he indirectly conveying product of 2 un chosen numbers.

For example, product of all numbers in set = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 and if Zach tell product 1 x 2 x 4 x 5 x 7 = 280 then obviously the product of numbers that he hasn't chosen is 5040/280 = 18 = 3 x 6.

Now, there are only 2 products viz. 12 and 6 which have more than 1 pair of numbers. 

The product 12 can be from pairs - (3,4) and (6,2)

The product 6 can be from pairs - (1,6) and (2,3) 

Here if the sum of un chosen numbers is odd (or even) then sum of other 5 chosen numbers also must be odd (or even).

In above cases, 6 has pairs whose sum is odd always and hence sum of other 5 numbers would be odd. In that case, Claudia would have been sure with if sum of numbers selected by Zach is either odd or even.

While in other case, the product 12 has 1 pair having sum odd (3,4) and other pair having sum even (6,2). Hence, Zach must have 'indirectly' suggested product 12 as a product of un chosen numbers that's why Claudia is saying that she doesn't know if the sum of numbers selected by Zach is even or odd.

Hence, the product of numbers selected by Zach = 5040 / 12 = 420.    

 

A Verbal Arithmetic Puzzle

In this equation, each of the letters represents uniquely a different digit in base 10:

YE × ME = TTT.

What is E + M + T + Y?

Here is solution!

A Verbal Arithmetic Puzzle - Solution

What was the puzzle?

Let's recall the original equation.

YE × ME = TTT.

Obviously, answer TTT is product of T x 111 = T x 3 x 37. 

So YE/ME = 37 and T x 3. Either way, E = 7.

And other number T x 3 is a 2-digit number with 7 at it's unit's place.

So that has to be 9 x 3 = 27.

Therefore,

YE (or ME ) = 37 and ME (or YE) = 27

Hence,
 
E + M + T + Y = 7 + 3 + 9 + 2 = 27.

The Number Game!

Let’s play a game. You name an integer from 1 to 10. Then we’ll take turns adding an integer from 1 to 10 to the number our opponent has just named, giving the resulting sum as our answer. Whoever reaches 100 first is the winner.

You go first. What number should you choose?

This is how you can be winner!

Winning The Number Game!

What was the game?

Here the player whose number 'forces' sum to fall in range of 90-99 will be ending on losing side. 

That means, somehow if you 'force' the total at some point to 89 then opponent has to fall in the range of 90-99 with his number. 

To get 'door' to total 89 you have to force the previous sum to 78 so that opponent is forced to open a 'door' for total 89 for you.

And so on backward you have to make stops at 67, 56, 45, 34, 23, 12, 1.

So you have to start with 1 & achieve all above milestones.

Let's verify our conclusion. Suppose you started with 1.

You       Sum      Opponent    Sum

1            -             8              9
3           12            5              17
6           23            7              30
4           34            9              43
2           45            4              49
7           56            10            66
1           67            3              70
8           78            2              80
9           89            4              93
7           100           

YOU WIN!

CheckMate The Opponent in 2 Moves!

Can White checkmate Black in two moves from this position?

Here are those 2 moves! 

2 Moves to CheckMate the Opponent

Where the game stands?


1.Move Rook to b3 i.e. perform Rb3.

2.That will force opponent to move king to a5 i.e. execute Ka5.

3. Finally,now move Rook to a3 (Ra3) and Checkmate the opponent.

Case of 3 Identical Notebooks

Three people all set down their identical notebooks on a table. On the way out, they each randomly pick up one of the notebooks. What is the probability that none of the three people pick up the notebook that they started with?

That's correct probability!

Probability in Case of 3 Identical Notebooks

What was the case?

Let's name peoples as Person - 1, Person - 2, Person - 3 and their notebooks as Notebook - 1, Notebook - 2 and Notebook 3 respectively.

Now there can be 6 ways 3 notebooks can be distributed among 3 persons like below.

(Here, for convenience, 3 different colors are assigned to the notebooks of 3 persons.)

As we can see, there are only 2 cases, where the each of person not getting own notebook. In rest of cases, at least 1 person got own notebook & notebooks of others are shuffled between 2.

So the probability that none of the three people pick up the notebook that they started with is 2/6 = 1/3. 

Unfair Game of Strange dice

Katherine and Zyan are playing a game using strange dice. Each die is a cube with six sides. Katherine's die has sides numbered 3, 3, 3, 3, 3, and 6. Zyan's die has sides numbered 2, 2, 2, 5, 5, and 5.

To play the game, Katherine and Zyan roll their dice at the same time and whoever rolls the higher value wins. If they play many times, who will win more frequently, Katherine or Zyan?

This person will be winning more! 
 

Advantage in Unfair Game of Strange Dice

What was the game?

Shortest Way : 

Katherine's die has sides numbered 3, 3, 3, 3, 3, and 6. And Zyan's die has sides numbered 2, 2, 2, 5, 5, and 5. That means whenever Zyan rolls a 2 then Katherine will always win. The probability that Zyan rolls 2 is 3/6 = 1/2. So in half of the cases, the Katherine will be winner of the game. Moreover, whenever, Zyan rolls a 5, Katherine can win if she rolls a 6. In short, in more than half cases, Katherine will win hence she has more advantage in this game.


Precise Way : 

Katherine will win if

1. Zyan rolls a 2 with probability 3/6 = 1/2

2. Zyan rolls a 5 (probability 3/6 = 1/2) and Katherine rolls a 6 (probability 1/6). So the probability for this win = 1/12.

Hence, Katherine has got 1/2 + 1/12 = 7/12 chances of winning.

Zyan will win if he rolls a 5 (probability 1/2) and Katherine rolls a 3 (probability 5/6) with probability = 1/2 x 5/6 = 5/12.  

That proves, Katherine has more chances (7/12 vs 5/12) of winning this game. 

Another Way : 

There can be 36 possible cases of numbers of cubes out of which in 15 cases Zyan seems to be winning and in 21 cases, Katherine is winning. Again, Katherine having more chances of winning (21/36 = 7/12) than Zyan (15/36 = 5/12). 

Search Number for '?'

Can you find the number for the '?' ?

Skip to the answer! 

In The Search of Number for '?'

Here is the question! 

First two equations clearly indicates it must be straightforward a + b equation.

Now if you observe closely all the equations, then you can find that the number never exceeds 8. So there are chances that all these number are in octal number system i.e. base 8 numbers.

Hence 11 in third equation is actually 8 in decimal. That's why 7 + 8 = 15 where decimal 15 is converted to octal 17. 

Verifying third equation as 15 + 16 = 31 in decimal is actually 17 + 20 = 37.

On the similar note, we have fourth equation in decimal as 31 + 32 = 63 which can be converted to octal as 37 + 40 = 77.

Hence the search for number in place of '?' ends with the number 77.

Profit Or Loss Or No-Profit No-Loss?

A man buys a horse for $60. He sells the horse for $70. He then buys the horse back for $80. And he sells the horse again for $90. How much profit did he make or did he loose in transaction? Or did he break even?

Confused? Find the right answer here!

And He Earned A Profit!

But what was the deal?

The confusion starts when he he buys same horse again. But if we look at it as two different transactions then it pretty straightforward.

At first he buys horse for $60 & sells it for $70. Here, he makes profit of $10. This is one transaction.

In next transaction, he buys same horse for $80 & sell the same for $90. Again, here he makes profit of $10.

In this way, the total profit he earns from these transactions is of $20.  

Difficult Puzzle of Sum and Product

Sum Sam and Product Pete are in class when their teacher gives Sam the Sum of two numbers and Pete the product of the same two numbers (these numbers are greater than or equal to 2). They must figure out the two numbers.

Sam: I don’t know what the numbers are Pete.

Pete: I knew you didn’t know the numbers… But neither do I.

Sam: In that case, I do know the numbers.

What are the numbers?

Want to know those numbers?

Solution - Difficult Puzzle of Sum and Product

Wait, read the puzzle once! 

Let's remind that the numbers are greater than or equal to 2; means those can't be either 0 or 1.

Now take a look at what Sam & Pete says - 

Sam: I don’t know what the numbers are Pete.

Pete: I knew you didn’t know the numbers… But neither do I.

Sam: In that case, I do know the numbers.

If Sam was told 4 then straightway he would have numbers 2,2 in mind as 3,1 combination is invalid.

If teacher had told Sam 5 as a sum then too Sam had correct pair of numbers 2,3 immediately as 4,1 or 5,0 are invalid combinations. 

So Sam must have at least number 6. Valid combinations for this sum are (2,4), (3,3).

If it was (3,3) then Pete would had 9 & he would have identified this combination correctly as (9,1) is not valid combination. Since he too didn't know exact numbers, it must be some different combination.

And if teacher had told Pete 8 then too he would have easily figured out correct combination of (2,4) as (8,1) is not valid.

So Pete can't have number 1,2,3,4,5,6,7, 8 or 9 or 11.

Now if he had 10 then only possible combination (2,5) and he would have that immediately. So he wouldn't have made the statement that he too didn't know numbers.

Let's assume that he had number 12 as product. Now in this case valid combinations are (2,6), (3,4).  The sums of these valid combinations are 8 & 7 respectively.

Now depending on what sum the Sam had; he can identify the correct pair of numbers easily.

Even/Even x Even - Odd ?

Can an even number, divided by another even number, times another even number ever equal an odd number?

The three even numbers can be different numbers.



Did you too think same? 

Even/Even X Even = Odd!

What was the question?

Absolutely, it's possible! Take a look at the examples below.

Totally depends on the what result of Even/Even comes out & what is the Even number that is being multiplied with the result.