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$6bcde$ is a five-digit number. If $6-d=x, c=y$ and $xy$ is a two digit square number in which each digit is a square number. If $b = p$, $e = q$ and $pq$ is a two-digit perfect square number in which each digit cube, the cube root of $6bcde$ is?

This question was on an exam I had taken about a month ago. Even now, I can't find the answer to this.

Prithvish Baidya
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1 Answers1

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If $xy$ is a two-digit square number in which each number is a square, then $xy = 49$; this is the only such number. So $x = 4$ and $y = 9$. Similarly, the only two-digit square number in which each digit is a cube is 81, so $p = 8$ and $q = 1$. All together, then, the number $6bcde$ is 68921, and $\sqrt[3]{68921} = 41.$

adjan
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Michael Seifert
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  • I thought $xy$ to be $x*y$. Don't you think information regarding this should have been given in the paper? – Prithvish Baidya Oct 22 '16 at 20:50
  • @PrithvishBaidya: I thought that too, at first. But if $x*y = 49$, then you must have $x = y = 7$. This means that $d = -1$, which is impossible. So I looked for another interpretation, and it worked. – Michael Seifert Oct 23 '16 at 13:14