# 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
3 weeks, 6 days ago

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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

- 3 weeks, 6 days ago

WOW thanks

- 3 weeks, 3 days ago

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.

- 3 weeks, 5 days ago

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).

- 3 weeks, 5 days ago

How did you conclude that there are 4 solutions?

- 3 weeks, 5 days ago

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.

- 3 weeks, 5 days ago

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

- 3 weeks, 5 days ago

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.

- 3 weeks, 5 days ago

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

- 3 weeks, 5 days ago

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.

- 3 weeks, 5 days ago

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

- 3 weeks, 5 days ago