PDA

View Full Version : '10x + y = 53'; What theory lets x=5 and y=3?

wdonuts
12-13-2009, 02:03 PM
I have the answer to an algebra equation that is the solution to a card
trick, but I do not understand the theory that allows the final step of the
solution to be valid.

Someone selects two cards and given the cards are '5' and '3,' the
following equation can be reached:
'10x + y = 53'
Dividing 53 by 10, the quotient is 5 and the remainder is 3, but I do not
understand what theory would make this step valid. I have considered simple
division and mod-ing, but I still don't understand what theory or property
makes this possible.

With the equation:
'12x + y = 38,'
dividing 38 by 12, the quotient is 3 and the remainder is 2, which are the
values of the two cards selected in another manipulation of the same card
trick. I believe this has something to do with varying the base of the
equation, but I'm still unsure of what makes this valid or if it is even
correct.

Thank you for any help.

Loren
12-13-2009, 02:39 PM
I'm not sure I understand your question. Here are my thoughts.
The value of any two digit number (in the decimal number system) is determined by multiplying the first digit (the ten's digit) by ten and the second digit (the unit's digit) by one, and adding those results together. The same principle extended applies to a three digit number, a four digit number, etc.
For instance the number 472 has a value of 4·100 + 7·10 + 1·2.
The number 38 has the value of 3·10 + 8·1. I don't know where your "12" comes from.

DrMike
12-14-2009, 08:39 PM
Someone selects two cards and given the cards are '5' and '3,' the
following equation can be reached:
'10x + y = 53'
Dividing 53 by 10, the quotient is 5 and the remainder is 3, but I do not
understand what theory would make this step valid.

In general, the equation 10x+y = 53 has many solutions. However if you know something about the values of x and y (eg, because you know the rules of the 'trick'), many of these solutions might be rules out. Perhaps you are left with only one solution.

For example, perhaps you know that both x and y are integers. Then, you can tell that y mod 10 is 3 (that is, the remainder when you divide y by 10 is 3), and that x = (53-y)/10.

If you also know that x and y must be in the range 0 to 9, then you know that y=3, and x=(53-y)/10 = 50/10 = 5.

If you had some other number T instead of 53, you could prove in general that "If T is an integer between 0 and 99, and if x and y are integers between 0 and 9, and 10x+y=T, then x and y are respectively the quotient and remainder when you divide T by 10" however, I'm not going to give the proof in detail here...