The King and the Mathematician

There are 64 squares. In the first square the king puts $1, in the second he puts $2, in the third he puts $4, and so on. In general, in the nth square he puts $2^{n-1}$ dollars. In the last square which is 63rd square, he puts $2^{64-1} = 9,223,372,036,854,775,808$ dollars.

And the sum of the sequence $1, 2, 4, 8, 16, 32, ..., 2^{n-1} = 2^{n} - 1$ , so overall the king would be putting in $2^{64}-1 = \$18,446,744,073,709,551,615$ dollars, if he had that kind of money that is (this amount is much higher than the total worth of the kingdom which is 1500 trillion dollars).

Hence the king will go bankrupt way sooner than reaching the last square and the man will be the new king.

The first row has '8' boxes and the last box contain '128' so that lest '7' boxes will be multiply with '128' '7' more times....like ( 128 X 128 X 128 X 128 X 128 X 128 X 128 X 128 = 7.2057594E16)

We can make the value of the first box =2^0, since the values are in multiples of 2. Write them as exponents to make it simple. Since 2^0=1, 2^1=2 and so on, if we number the boxes horizontally from 0-7, and vertically from 1-8 (for each row), we get in the box in the bottom right corner to be 2^(7*8)=2^56=7.205e16, which is much more than the fortune of the king. the king can't give the mathematician this amount, since he doesn't have it, he'd go bankrupt.

just watch out the figures and the order the figures are in and you will get the answer. just get the logic. and become a king

Very easy. There are 64 squares. Starting from 2^0 from the start it goes to 2^63, adding all the values in between i.e. 2^0 + 2^1 + 2^2 +................ + 2^63 = 2^64 -1 . Now , 2^10 is 1024 which is nearly one thousand or three 0 s. So, 2^ 60 will have 3 x 6 = 24 zeroes . One trillion is only 12 zeroes. So the value of Gold here, becomes, much more than the worth of King`s Kingdom. So, the King becomes bankrupt much before reaching the last square and the mathematician becomes the King :)

there are 64 squares, the first square is 2^0, next is 2^1, so the last square is 2^23. sum of 64 squares is (2^n)-1, is larger than 1500 trillion dollars

This is really easy problem. The mathematician will get: 1 + 2 + 2^2 + 2^3 + ... + 2^63 = 18 446 744 073 709 551 615 dollars

With the 64 squares grid,putting double tthe dollars of the previous one 64th square will have 2raise to power 64-1 that is Us $ 9,223,372,036,854,775,808 when we add up all 64 squares filled with mentioned do;;ars sum will be $ 18,446,744,073,709,551,615 which is huge amount more than the worth of kingdom and king will be bankrupt. K.K.GARG,India

Its logical. No need of solving. See , if the man mathematician wants the whole kingdom desperately, he will do anything to get it, even by tricking the king.

Easy to see ,without calculating what might be confused, by considering the last square at the bottom right. According to the figure, there are 64 small squares; beginning at the top left with 2^0, it then exactly end with 2^63 = (2^60)x(2^3) = (1024^6)x8. So, the king cost about a QUINTILLION dollars on the final time.

guys , i am confuse....by one thing...

1500 trillion = 1500 x 10^(13) = 1.5x10^(16)

i press calculator 2^63 = 9.22x10^(18)

i calculate the summation of the geometric progression a , ar^2 , ar^3 ,........,ar^63 , a=1 , r=2

a(1-r^n)/(1-r)

= 1 [1-2^(63)] / (1-2) = 9.22 x10^(18) = 2^63

i find it awfully weird , then i do summation manually 2^53 + 2^62 +...+ 1 = 2^64

WHY???????????i thought my summation in geometric sequence is correct but i got different answer when i sum manually......

Hello,

as for 64 squares of having a=1 , r = 2,by geometric,

Sum of the 1st 64 terms = 1( 2^64 -1) / ( 2-1) ~ 1.84467 x 10^19 ,

By the 1st place, kingdom of the king = 1.5 x 10^15 = 1500 Trillions...

1.84467 x 10^9 > 1.5 x 10^15 by ~ 12 300 times....haha...sure the King became a beggar onwards...