For a positive integer n, two payers a and b play the following game: given a pile of s stones, the players take turn alternatively with a going first. on each turn the player is allowed to take either one stone, or a prime number of stones, or a positive multiple of n stones. the winner is the one who takes the last stone. assuming both a and b play perfectly, for how many values of s the player a cannot win?

the aswere is the one with all 8's and the second one is with the 10,8 and 6

step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)

4.28095238095 that is the answer