MHB Determine how many licence plates would cost $100

  • Thread starter Thread starter WMDhamnekar
  • Start date Start date
  • Tags Tags
    Plates
AI Thread Summary
Each vehicle license plate consists of three letters followed by three digits, with costs varying based on the letters' positions in the alphabet and the digits used. The cost for digits is $n for n>0 and $10 for 0, while letters cost $1 to $26 based on their alphabetical order. The maximum cost for a license plate is $108, represented by "ZZZ 000." To find how many license plates total $100, the problem translates to distributing 8 indistinguishable items into 6 distinguishable categories, resulting in 1287 possible combinations. Thus, 1287 license plates can be produced for $100.
WMDhamnekar
MHB
Messages
376
Reaction score
28
In a certain state of a certain country, each vehicle license plates have exactly three letters followed by three digits. We are told that to produce such a license plate, it costs $n for each digit n>0 and $10 for each digit 0. For letters the cost is proportional to the position of the letter in the alphabet, namely \$1 for A, \$2for B, so on and so forth, upto $26 for Z.

Now, how to determine how many license plates would cost $100?

Answer:- I don't understand how to answer this question. I think linear programming will help here.
 
Last edited:
Mathematics news on Phys.org
This question seems to be difficult and maybe linear programming can help here.
 
Dhamnekar Winod said:
In a certain state of a certain country, each vehicle license plates have exactly three letters followed by three digits. We are told that to produce such a license plate, it costs $n for each digit n>0 and $10 for each digit 0. For letters the cost is proportional to the position of the letter in the alphabet, namely \$1 for A, \$2for B, so on and so forth, upto $26 for Z.

Now, how to determine how many license plates would cost $100?

Answer:- I don't understand how to answer this question. I think linear programming will help here.
Hello,
After working on finding out the answer to this question, eventually i suceeded. The answer to this question is 1287 license plates would cost \$100.

Justification to the answer:-

The maximum possible cost is \$108 of the license plate having ZZZ 000. Finding the number of license plates having the cost of \$100 is equivalent to finding how many ways 8 indistinguishable balls can be put into 6 distinguishable cells. So $\binom{n+r-1=13}{r=8}=1287 $ license plates.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top