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1) John noticed that the amount he was paying for his lunch was a rearrangement of the digits of the amount of money he had in his pocket, and that the money he had left over was yet another rearrangement of the same three digits! How much money did John start with?

2) A small number of cards has been lost from a complete pack. If I deal among four people, three cards remain. If I deal among three people, two remain and if I deal among five people, two cards remain. How many cards are there?

User Makim
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1 Answer

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There doesn't seem to be enough information to answer problem 1.

I would ask your teacher for clarification.

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Problem 2

Answer: 47 cards

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Step-by-step explanation

x = number of cards, which is some positive whole number.

Your teacher has provided these facts:

  1. If I deal among four people, three cards remain.
  2. If I deal among three people, two cards remain.
  3. If I deal among five people, two cards remain.

which I'll call "fact 1", "fact 2", "fact 3" in that order.

For now I'll focus on fact 3.

Recall that multiples of 5 either end with 0 or 5. Example: 25 is a multiple of 5, and so is 60.

Fact 3 says that we get a remainder 2 when dividing by 5. Therefore, x must be something like 25+2 = 27 or 60+2 = 62. I added 2 to each previous example mentioned earlier.

In short, the value of x must end with 2 or 7.

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Now focus on fact 1.

The multiples of 4 are {4, 8, 12, ...} all of which are even since 2 is a factor of each.

Add remainder 3 to each item to get {7, 11, 15, ...} and it should be fairly obvious that all of these items are odd.

What does this mean? It means we know 100% that x ends with 7 in the units digit. If x has 2 in the units digit, then x would be even, but it's not in the set {7, 11, 15, ...}

So this is very useful info that x ends with 7.

x could be any of these: {7, 17, 27, 37, ...}

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We've looked at facts 1 and 3 so far. The only thing left is fact 2.

List the multiples of 3 to get

{3, 6, 9, 12, 15, 18, ...}

then add on the remainder 2 giving this new updated set

{5, 8, 11, 14, 17, 20, ...}

We see that 17 is in that set which fits the criteria of "the number ends with 7". So it's possible that 17 is the answer.

But 17/4 = 4 remainder 1 which doesn't fit fact 1 (since we want remainder 3 instead).

But we can add on multiples of 3*5 = 15 to extend to this new set of

{17, 32, 47, 62, 77, ...}

then erase stuff that ends with 2 to get

{17, 47, 77, ...}

Through guess and check, 47 will work out as shown in the next section below. Therefore, 47 is the final answer.

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Check:

  • 47/4 = 11 remainder 3
  • 47/3 = 15 remainder 2
  • 47/5 = 9 remainder 2

We get the proper remainders mentioned earlier, so the answer has been confirmed.

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Side note: The convention for problems like this is to use the Chinese Remainder Theorem. However, I decided not to use it because I wasn't sure if your teacher has covered modulo arithmetic yet or not. If s/he has covered it, then I recommend using that method instead of what I posted above. It's a much more efficient option.

User Jingles
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