One day, I had a number of dollars in my wallet. I went into a cake shop and saw this special limited edition mousse cake. I went to an employee in the shop and had a conversation with him. The following conversation ensued:
Me : Hello there! May I ask how much that mousse cake is?
Employee : Oh, it's ...... dollars per slice.
I checked my wallet to see how much money I had.
Me : Oops, I can only buy 4 of them.
Employee : That's a pity! The mousse cake is really good. Fine, I lower the price by 2 dollars just for you!
Me : Wow thanks! Now I will be able to buy at most 5 of them.
Employee : I think you should buy more. I lower the price again by 3 dollars just for you, okay?
Me : You're so nice! I will be able to buy at most 6 of them!
Employee : Just nice! Our mousse cake is always cut into 6 slices. I will prepare the mousse cake for you!
Note: The "......" part is blanked out for you to solve this problem.
If the amount of money I had in my wallet and the amount per slice of the mousse cake are both positive integers, find the smallest amount that I had in my wallet.
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Let X be the amount I have in my wallet. Let Y be the price per slice.
So 4Y "< or =" X < 5Y,
5(Y - 2) "< or = X" < 6(Y - 2), and
6(Y - 5) "< or =" X < 7(Y - 5).
We want 5Y to be more than 5(Y - 2) and 6(Y - 5), 6(Y - 2) to be more than 6(Y - 5) and 4Y, and 7(Y - 5) to be more than 4Y and 5(Y -2).
5Y > 5Y - 10 is definitely correct. 5Y > 6Y - 30, 30 > Y.
6Y - 12 > 6Y - 30 is definitely correct too. 6Y - 12 > 4Y, 2Y > 12, Y > 6.
7Y - 35 > 4Y, 3Y > 35, Y > "11.6666666.....". 7Y - 35 > 5Y - 10, 2Y > 25, Y > 12.5
So let's join up all the inequalities related to Y:
12.5 < Y < 30
The least possible positive integer for Y is 13, so let's go back to the linear equations:
52 "< or =" X < 65,
55 "< or =" X < 66, and
48 "< or =" X < 56.
The only possible positive integer for X is 55.