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math

Replace each letter with a number to make the problem correct. Identicle letters are placed with identicle numbers. PARTS x 4 = STRAP

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Dylan

  1. PARTS = 21978

    I got it the hard way, by trial and error, after realizing that P had to be 2 and S had to be 8.

    This kind of problem is not algebra; it is a riddle. There is no guarantee that there would be only one solution. It just turns out that way.

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    drwls

  2. PARTS x 4 = 87912

    In looking at the numbers that I tried, without using the same number twice, only those with P = 2, A = 1, S = 8 and T = 7 would start out with 8 and end with 12 when multiplied by 4. That left only a few R numbers to try.

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    drwls

  4. WIth different letters but the same result:

    ABCDE x 4 = EDCBA

    Lets take it one number at a time.
    1--A must be either 1 or 2 or the answer would be a 7 digit number.
    2--If A = 1, the 4 x E has to end with 1 which cannot be, so A = 2.
    3--If A = 2, then E can only be 3 or 8 for the product to end in 2.
    4--But since 4 x E = E, E must be 8 to satisfy both conditions.
    5--4 x B can have no carryover to 4 x 2 so B must be 1 as A is already 2.
    6--4 x D + the 3 carryover must equal 11, 21, or 31 since B = 1.
    7--So D can only be 2 or 7 but since A = 2, D must be 7.
    8--4 x 1 must have a carryover of 3 to become 7 so C must be 7, 8 or 9.
    9--Since D = 7 and E = 8, C must be 9.
    10--Therefore, the numbers are 21978 x 4 = 87912.

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    tchrwill

  5. 5--4 x B can have no carryover to 4 x 2 so B must be 1 as A is already 2.

    why can't B be 0 at step 5? At this point, B can be 0,1,2, and you can eliminate 2 per your reason, but you can't eliminate 0 yet--which makes the rest of your steps invalid.

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    question

  6. 4A72
    B85C
    92D6
    Answer
    E 7775

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    Anonymous

  8. thanks you are a big help

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    marcus

  9. I NOT HAS MINE

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    JEFF

  10. IT IS HAS MINJES

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    JEFF

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