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Given that equation $xyzp=4\cdot pzyx$ is valid, how do I solve showing my work clearly? By trial and error the valid answers are $x=8,y=7,z=1,p=2$. These are the only conditions provided.

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Is everything needed in your question? –  Amzoti Jul 7 '13 at 4:35
    
Does your question mean, "how to solve the problem without trial and error?" –  フラン Jul 7 '13 at 4:38
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I suspect that the variables are supposed to be digits (not positive reals), and we are talking about concatentation rather than multiplication. –  Gerry Myerson Jul 7 '13 at 4:41
    
@Gerry: Which makes the $4$ even more confusing... –  Zev Chonoles Jul 7 '13 at 4:42
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This question appears to be off-topic because "If and only if the (whys) are answered because here take it I never gave out my answers." –  Will Jagy Jul 7 '13 at 5:14

2 Answers 2

Here's a start (assuming I have the right interpretation of the problem in the comments).

  1. $p$ must be even (why?)

  2. $p\lt3$ (why?)

  3. So $x$ is 8 or 9 (why?)

  4. But $x$ can't be 9 (why?)

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If and only if the (whys) are answered because here take it I never gave out my answers. –  NGILAZI BANDA JOSHUA Jul 7 '13 at 4:57
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I'm sorry --- I have no idea what you mean by that. –  Gerry Myerson Jul 7 '13 at 5:57

p z y x

× 4

x y z p x × 4 = p in unit place. similarly p × 4 = x

4 digit number × 4 gives 4 digit answer, p < 3 But p = 4 × x is even so p = 2 and x = 8 so it becomes 2 z y 8

× 4

8 y z 2 it shows that there is no carry forward in z × 4 so z must be 1 2 1 y 8

× 4

8 y 1 2 so you can calsukate y 8 × 4 = 32 ----- 3 c/f. y × 4 + 3 = 1 in unit place so y = 7

x = 8,y = 7,z = 1,p = 2

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