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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!
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.
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!
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!
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.
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!