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## Question

There were six prices for various TV sets sold at the store: \$231, \$273, \$429, \$600.60, \$1001, and \$1501.50. One day, a motel owner came in and bought a bunch of TVs. The total came to \$13519.90 but the bill of sale was lost. How many of each TV type did the motel guy buy?

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No multiple of .60 ends in .90. So an odd number a of the \$1501.50 sets, along with b = 4 (mod 6) of the \$600.60 sets were purchased. Along with some number of the other sets. There are only a few possible combinations of (a, b) that come in under the total. I'll figure out the other sets later today.

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I have been working on this a few days.

bonanova is right

There must be
- an odd number of the \$1501.5 TVs and
- 5n + 4 of the \$600.6 models [for n = 0, 1, ...]

interestingly these 2 prices have a 5/2 ratio, so the combinations have the same values

e.g., (4* 600.6) + 3*(1501.5) = (9*600.6)+ (1*1501.5)

If we make the assumption that there is a unique solution, then it is probably safe to work with the only unique value to these two sets

(4*600.6) + (1501.5) = 3903.90

I have not found a good metholoology for finding combination of the other 4 sets to equal 9616.0.

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(12*231)+(9*273)+(5*429)+(4*660.60)+(1*1501.50)+(2*1001)=13519.90

Edited by harey
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Question for harey: how did you solve for the quantitites??

Thanks ....

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Set aside the sets calculated by Bonanova:
13,519.90-(4*660.60)-(1*1501.50)=9,376

As 600.60*5=1501.50*2=3003=3*1001, remain TVs for 231, 273, 429, 1001 (multiple solutions reserved).

231=3*7*11
273=3*7*13
429=3*11*13
-> all divisible by 3

9+3+7+6=25; so 9,376 is not divisible by 3;
-> there are at most 8 TVs priced 1,001 and only 7,374 and 1,368 fits.

The same way substract multiples of 231 and consider multiples of 39 (3*13, common divisor of 273 and 429).

Alternatively, I think the problem could be solved by Diophantine equations, but I never studied them.

Edited by harey
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Very nice apprach!

I missed that relationship  ... good job.

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Interesting.  I found a different solution and I thought my solution was unique.

I thought my series was unique, too. ##### Share on other sites

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Let a, b, c, d, e, and f be the number of TV's bought at the prices in their listed order. Thus, after
rewriting the prices, 3003a/13 + 3003b/11 + 3003c/7 + 3003d/5 + 3003e/3 + 3003f/2 = 135199/10
Multiplying by 3003 gives a/13 + b/11 + c/7 + d/5 + e/3 + f/2 = 135199/30030 = 135199/(2*3*5*7*11*13).
Thus, by clearing fractions, 2310a + 2730b + 4290c + 6006d + 10010e + 15015f = 135199.
Now, consider the equation mod 13, 11, 7, 5, 3, and 2. Each element on the left of the equation
is nonzero mod 13, 11, 7, 5, 3 and 2, respectively, and is zero modulo each of the other primes.
Thus, the equation yields the following 6 equations: .9a = 12 = 9*10 mod 13,
2b = 9 = 2*10 mod 11,. 6c = 1 = 6*6 mod 7, ..d = 4 mod 5, . 2e = 1 = 2*2 mod 3, . f = 1 mod 2.
Thus, a >= 10, b >= 10, c >= 6, d >= 4, e >= 2, and f >= 1. (... all these can be more strictly defined ...)
However, 231*10 + 273*10 + 429*6 + 600.60*4 + 1001*2 + 1501.50*1 = \$13519.90.
Therefore, ( 10, 10, 6, 4, 2, 1 ) is the only solution.

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I calculated with 660.60 and the price is 600.60.

Just a small typo as I often do...

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