Home / Assignment Help / Yi and Sue play a game. They start with the number 42000. Yi divides by a prime number, then passes the quotient to Sue. Then Sue divides this quotient by a prime number and passes the result back to Yi, and they continue taking turns in this way. For example, Yi could start by dividing 42000 by 3. In this case, he would pass Sue the number 14000. Then Sue could divide by 7 and pass Yi the number 2000, and so on. The players are not allowed to produce a quotient that isn’t an integer. Eventually, someone is forced to produce a quotient of 1, and that player loses. For example, if a player receives the number 3, then the only prime number (s)he can possibly divide by is 3, and this forces that player to lose. Who must win this game

Yi and Sue play a game. They start with the number 42000. Yi divides by a prime number, then passes the quotient to Sue. Then Sue divides this quotient by a prime number and passes the result back to Yi, and they continue taking turns in this way. For example, Yi could start by dividing 42000 by 3. In this case, he would pass Sue the number 14000. Then Sue could divide by 7 and pass Yi the number 2000, and so on. The players are not allowed to produce a quotient that isn’t an integer. Eventually, someone is forced to produce a quotient of 1, and that player loses. For example, if a player receives the number 3, then the only prime number (s)he can possibly divide by is 3, and this forces that player to lose. Who must win this game

Naturally, any integer n larger than 127 will return 127equiv127mod n, and of course 127equiv0mod127, so we restrict the possible solutions to 1le n<127.

Now,

127equiv7mod n

is the same as saying there exists some integer k such that

127=nk+7

We have

implies 120=nk

which means that any n that satisfies the modular equivalence must be a divisor of 120, of which there are 16: {1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120}.

In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,

127=21cdot6+1iff127equiv1equiv7mod6

(If we’re allowing n=1, then I see no reason we shouldn’t also allow 2, 3, 4, 5, 6.)

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