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
4 months, 2 weeks 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)\mod {(10)}. There are only 4 entries of 1

Mahdi Raza - 4 months, 2 weeks ago

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WOW thanks

Vicky Clearwater - 4 months, 2 weeks ago

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We must solve xy1xy \equiv 1 mod 10.

For any xx 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 xx, there is no solution. Therefore there are ϕ(10)=4\phi(10) = 4 solutions.

Arjen Vreugdenhil - 4 months, 2 weeks 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).

Arjen Vreugdenhil - 4 months, 2 weeks ago

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How did you conclude that there are 4 solutions?

Mahdi Raza - 4 months, 2 weeks ago

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

The Euler totient function ϕ(10)=ϕ(2)ϕ(5)=14=4\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.

Arjen Vreugdenhil - 4 months, 2 weeks ago

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

Mahdi Raza - 4 months, 2 weeks ago

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@Mahdi Raza No. Since 2 is a zero divisor modulo 10 (2×502\times 5 \equiv 0), it is not an invertible element. You would reduce the equation to xy1   mod 5,x'y' \equiv 1 \ \ \ \text{mod}\ 5, which has ϕ(5)=4\phi(5) = 4 solutions. Afterward, double either xx or yy, and optionally add 5 to the other, resulting in 16 solutions. In the table below, the first column has the values of (x,y)(x',y'): (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)\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.

Arjen Vreugdenhil - 4 months, 2 weeks ago

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

Mahdi Raza - 4 months, 2 weeks ago

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@Arjen Vreugdenhil Note that the "doubles" are all of the form (x,y)=(2u,2v)(x,y) = (2u,2v), where uv123   mod 5.uv \equiv \frac12 \equiv 3\ \ \ \text{mod 5}. This equation has four solutions.

Arjen Vreugdenhil - 4 months, 2 weeks ago

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@Mahdi Raza But for the other non-zero divisors (xy1,3,7,9xy \equiv 1, 3, 7, 9 mod 10) there are four solutions each.

Arjen Vreugdenhil - 4 months, 2 weeks ago

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