Dividing question, possibly related to factor theorem

[tony24810]

Homework Statement

The total cost for 88 adult tickets was $_293_. Because the printing machine was not functioning well, the first and the last digits of the 5-digit number were missing. If the cost for each ticket is $P, where P is an integer, find the value of P.

Homework Equations

i think factor theorem should be used.

The Attempt at a Solution

attempt 1

cost = 10000a + 2930 + b
88P = 10000a + 2930 + b



attempt 2
try to make x = 100
and make divider 88 to become 100-12 = 88, thus divider becomes x-12

generate a general equation
f(x) = a(x^2) + 2930 + b

then because it can be wholly divided by x-12,
f(12) = 0
144a + 2930 + b = 0



i don't seem to be getting anywhere, please help

 

[verty]

Hint: if you know the remainder of 10000/88, this should be easier.

 

[rcgldr]

This is a division / remainder problem / puzzle. How many possible values are there for a? You could just try them all.

 

[tony24810]

Hint: if you know the remainder of 10000/88, this should be easier.

113*84 = 9944
the remainder is 56

ok let me try

 

[symbolipoint]

EDIT: See next post which comes after this one!

You can make better progress using "attempt 1". You realize that you want a FIVE digit number for the total sales, and that your "a" and your "b" must be NATURAL numbers LESS than 10. This problem seems not to be any factor or remainder theorem problem. You may simply test several combinations of a and b using natural numbers between 1 and 9 inclusive, until you find results that make sense.

 

[symbolipoint]

WAIT... this may be even easier if not simpler.

FIVE digit number, so you want to find P, the price per ticket. The highest amount of sales possible is 92939 dollars and the smallest amount of sales possible is 12930. You can get a range of solutions for P.

88*P=92939 for maximum P
88*P=12930 for minimum PSome test calculations show that these are expensive tickets. We can pick lowest possible ticket prices if our leading digit in _293_ is picked as 1. From there, calculations for the ones place can be included to see what values give "even money" for total sales.

 

[rcgldr]

I'm not sure knowing the remaind of 10000/88 = 56 helps here. There are only 10 cases to consider:

88 * P = 0293_
88 * P = 1293_
88 * P = 2293_
...
88 * P = 9293_

Only one of these will work out to exact integer values (assuming P is an integer).

 

[junaid314159]

88P = 10000a + 2930 + b

- Hint 1: Think of the numbers that divide the LHS. They must divide the RHS as well.
- Hint 2: Show that a + b = 7
- Hint 3: Show that b is even

This will narrow the search to only four possibilities.

Junaid Mansuri

 

[willem2]

88P = 10000a + 2930 + b

- Hint 1: Think of the numbers that divide the LHS. They must divide the RHS as well.
- Hint 2: Show that a + b = 7
- Hint 3: Show that b is even

This will narrow the search to only four possibilities.

Junaid Mansuri

There's only a single choice of b that will make the number divisible by 8, and you'll only have to consider the last 3 digits, because 1000 is divisible by 8.

 

[symbolipoint]

I'm not sure knowing the remaind of 10000/88 = 56 helps here. There are only 10 cases to consider:

88 * P = 0293_
88 * P = 1293_
88 * P = 2293_
...
88 * P = 9293_

Only one of these will work out to exact integer values (assuming P is an integer).

88P = 10000a + 2930 + b

- Hint 1: Think of the numbers that divide the LHS. They must divide the RHS as well.
- Hint 2: Show that a + b = 7
- Hint 3: Show that b is even

This will narrow the search to only four possibilities.

Junaid Mansuri

There's only a single choice of b that will make the number divisible by 8, and you'll only have to consider the last 3 digits, because 1000 is divisible by 8.

The cheapest solution is 12936 dollars, from a ticket price of $147.00.

 

[tony24810]

88P = 10000a + 2930 + b

- Hint 1: Think of the numbers that divide the LHS. They must divide the RHS as well.
- Hint 2: Show that a + b = 7
- Hint 3: Show that b is even

This will narrow the search to only four possibilities.

Junaid Mansuri

I don't understand why a+b=7

 

[tony24810]

There's only a single choice of b that will make the number divisible by 8, and you'll only have to consider the last 3 digits, because 1000 is divisible by 8.

This is great, the fact that 1000 is divisible by 8 is actually quite a helpful reminder, thanks!

 

[verty]

10000a + 2930 + b $\equiv$ 0 (mod 88)
(10000 % 88)a + (2930 % 88) + b $\equiv$ 0 (mod 88)
56a + 26 + b $\equiv$ 0 (mod 88)
56a $\equiv$ 62 - b (mod 88)
52 < 56a ≤ 62 (mod 88)

This is called modular arithmetic, calculating with remainders. The % sign here (read as "mod") means find the remainder. It's very useful but not often taught nowadays. My hint was to suggest using modular arithmetic.

 

[junaid314159]

@tony: The reason why a+b=7 can be seen by evaluating both sides of the second equation mod 11.