# The poker card Qs

From a stack of poker cards, the cards A, 2, 3, 4, 5, 6, 7, 8, 9, 10 of spades and A, 2, 3, 4, 5, 6, 7, 8, 9, 10 of Hearts were selected.

Out of these twenty cards, how many ways can one select two cards-one from Hearts and one from spades-such that the product of the two numbers leaves a remainder of 1 when divided by 10?

Note by Vicky Clearwater
1 year, 1 month ago

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## Comments

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A two-way table as such can be made. With each entry entered as product and then of $\mod {(10)}$. There are only 4 entries of 1

- 1 year, 1 month ago

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WOW thanks

- 1 year, 1 month ago

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We must solve $xy \equiv 1$ mod 10.

For any $x$ that is not a zero divisor modulo 10 (i.e. not a multiple of 2 or 5), this equation has a unique solution modulo 10. For other $x$, there is no solution. Therefore there are $\phi(10) = 4$ solutions.

- 1 year, 1 month ago

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Of course, this kind of quick argument only works if you know some abstract algebra.

If this is a high school assignment, you're best off reasoning through it in a different way (e.g. making a table like Mahdi did).

- 1 year, 1 month ago

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How did you conclude that there are 4 solutions?

- 1 year, 1 month ago

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Sorry, that's the part I forgot. I just edited my solution.

The Euler totient function $\phi(10) = \phi(2)\cdot \phi(5) = 1 \cdot 4 = 4$ gives the number of elements in the multiplicative group modulo 10, which is the number of non-zero divisors, as well as the number of invertible elements modulo 10.

- 1 year, 1 month ago

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Can this be concluded for $2 \mod{(10)}$ as well.

- 1 year, 1 month ago

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No. Since 2 is a zero divisor modulo 10 ($2\times 5 \equiv 0$), it is not an invertible element. You would reduce the equation to $x'y' \equiv 1 \ \ \ \text{mod}\ 5,$ which has $\phi(5) = 4$ solutions. Afterward, double either $x$ or $y$, and optionally add 5 to the other, resulting in 16 solutions. In the table below, the first column has the values of $(x',y')$: $\begin{array}{c|cccc} (1,1) & (2,1) & (1,2) & (2,6) & (6,2) \\ (2,3) & (4,3) & (2,6) & (4,8) & (7,6) \\ (3,2) & (6,2) & (3,4) & (6,7) & (8,4) \\ (4,4) & (8,4) & (4,8) & (8,9) & (9,8) \\ \end{array}$ There are some doubles here; the actual number of solutions is 12.

- 1 year, 1 month ago

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Yeah, I now understand. This is a good way actually. Thanks!

- 1 year, 1 month ago

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Note that the "doubles" are all of the form $(x,y) = (2u,2v)$, where $uv \equiv \frac12 \equiv 3\ \ \ \text{mod 5}.$ This equation has four solutions.

- 1 year, 1 month ago

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But for the other non-zero divisors ($xy \equiv 1, 3, 7, 9$ mod 10) there are four solutions each.

- 1 year, 1 month ago

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