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mmrproblem
Joined: 05 Nov 2004Posts: 15
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Posted: Fri Nov 26, 2004 4:50 am Post subject: digits
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Let xyz be a number with three digits.(base 10).Find all xyz
numbers that satisfy xyz=100x+10y+z=x^3+y^3+z^3
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Wilson
Frequent VisitorJoined: 20 Oct 2004Posts: 74
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mmrproblem
Joined: 05 Nov 2004Posts: 15
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Posted: Sat Nov 27, 2004 4:43 am Post subject: digits
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Thank you Wilson,but I don’t know Chineese.Any site in English?
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Curtiss20184
Frequent VisitorJoined: 30 Aug 2004Posts: 38
Location: HK F7
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Posted: Sat Nov 27, 2004 6:50 pm Post subject:
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I found one more solution for the 11digits .
It is 32164049651 ….it is quite easy to found if you have a solution end with 0 as the rightest digit. _________________
I am a Form 7 Student in Hong Kong.
Last edited by Curtiss20184 on Sat Nov 27, 2004 6:57 pm; edited 1 time in total
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Curtiss20184
Frequent VisitorJoined: 30 Aug 2004Posts: 38
Location: HK F7
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Posted: Sat Nov 27, 2004 6:57 pm Post subject:
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Curtiss20184 wrote:
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I found one more solution for the 11digits .
It is 32164049651 ….it is quite easy to found if you have a solution end with 0 as the rightest digit. |
_________________
I am a Form 7 Student in Hong Kong.
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