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Posers and Puzzles

Posers and Puzzles

  1. 29 Jul '04 18:23
    A customer at a 7-11 store selected four items to buy, and was told that the cost was $7.11. He was curious that the cost was the same as the store name, so he enquired as to how the figure was derived. The clerk said that he had simply multiplied the prices of the four individual items. The customer protested that the four prices should have been ADDED, not MULTIPLIED. The clerk said that that was OK with him, but, the result was still the same: exactly $7.11. What were the four prices?

    -Ray.
  2. 31 Jul '04 16:46
    It would seem to me that the answer to this problem will necessarily be those values that have exactly two decimal places. Or, if you prefer to view the price as 711 cents, then the prices will necessarily be integers. This limits the set of possible results.

    -Ray.
  3. Standard member TheMaster37
    Kupikupopo!
    04 Aug '04 17:56
    http://www.redhotpawn.com/board/showthread.php?id=11583
  4. 04 Aug '04 18:04
    Originally posted by TheMaster37
    http://www.redhotpawn.com/board/showthread.php?id=11583
    What were the odds of that?

    -Ray.
  5. Standard member TheMaster37
    Kupikupopo!
    04 Aug '04 18:47
    Originally posted by rgoudie
    What were the odds of that?

    -Ray.
    Very small, i just happened to remember that post, i do wish to find out a way to calculate it though.

    The best way i can think of is that you write the numbers as sum of powers of 10, then multiply is out, add it up, get conditions for the 15 parameters, then solve it...
  6. 04 Aug '04 19:16
    Originally posted by TheMaster37
    Very small, i just happened to remember that post, i do wish to find out a way to calculate it though.

    The best way i can think of is that you write the numbers as sum of powers of 10, then multiply is out, add it up, get conditions for the 15 parameters, then solve it...
    Sounds like a feasible plan.

    Given that the task was simply to discover the values that work, I would have come up some small program to find the values using brute force.

    -Ray.
  7. Standard member TheMaster37
    Kupikupopo!
    04 Aug '04 20:12
    Originally posted by rgoudie
    Sounds like a feasible plan.

    Given that the task was simply to discover the values that work, I would have come up some small program to find the values using brute force.

    -Ray.
    Bah! I bet you'd have the answer faster
  8. 07 Aug '04 13:06
    Originally posted by TheMaster37
    Bah! I bet you'd have the answer faster
    Challenge - first person to find a way to derive the answer with pencil and paper only wins.
  9. 07 Aug '04 22:57
    Originally posted by iamatiger
    Challenge - first person to find a way to derive the answer with pencil and paper only wins.
    Managed It (without looking at the answer)!

    a*b*c*d = 7.11
    a+b+c+d = 7.11

    We know that a,b,c and d can have at most 2 decimal places

    let A=100*a, B=100*b, C=100*c, D=100*d
    then A,B,C and D are integers, and

    A+B+C+D = 711 (we've multiplied both sides by 100)
    A*B*C*D = 711*10^6 (we've multiplied both sides by 10^8)

    next factorise 711*10^6
    711*10^6 = 79.2.2.2.2.2.2.3.3.5.5.5.5.5.5

    We then know that A, B C and D must be made up from some combination of those factors. And A + B + C + D = 711

    Its pretty easy to then try various possible combinations on paper, considering i) what we can multiply the largest factor of 79 by, without "busting" and ii) How a total ending in 1 can be obtained (this is quite a limiting consideration as numbers with a factor of 5 can only end in 0 or 5)

    After some fiddling I tried:
    79.2.2 + 5.5.5 + 5.5.2.2 + 5.3.3.2.2 = 316+125+100+180 = 721
    Hmm, close - Is there any way to reduce this total by 10? - well, if we swap a factor of 2 and a factor of 3 round in the last two numbers that should make the total slightly smaller....
    79.2.2 + 5.5.5 + 5.5.3.2 + 5.3.2.2.2 = 316 + 125 + 120 + 150 = 711!

    So the answer is $3.16, $1.50, $1.25 and $1.20