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Can you find the 10 digit number* which contains each of the digits 0-9* exactly once, and when divided by 9 the quotient is a perfect cube which has the digit 7 in it exactly twice*?

*in its base 10 representation

For clarification:

  • 1014107283 does not qualify, since although 4833=112678587 has two 7s, when multiplied by 9 the product (1014107283) is missing digits 5, 6 and 9.
  • 1963504872 does not qualify, since although 6023=218167208 and when multiplied by 9 the product 1963504872 is pandigital, 218167208 has only one 7.
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It is

1905378624

which, divided by 9 gives

211708736

which contains two sevens and is the third power of

596

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