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Phil_C
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#1
Who am I? I am three digits in length, a perfect square, my fourth root is Prime as is the sum of my digits.
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kyuuki
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#2
Juwel
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#3
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#3
(Original post by Phil_C)
Who am I? I am three digits in length, a perfect square, my fourth root is Prime as is the sum of my digits.
Who am I? I am three digits in length, a perfect square, my fourth root is Prime as is the sum of my digits.
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lou p lou
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#4
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#4
(Original post by Phil_C)
Who am I? I am three digits in length, a perfect square, my fourth root is Prime as is the sum of my digits.
Who am I? I am three digits in length, a perfect square, my fourth root is Prime as is the sum of my digits.
lou xx
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GH
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#5
Juwel
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#6
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#6
(Original post by 2776)
I agree with 625, how did u work that out?
I agree with 625, how did u work that out?
2^4 is not 3 digits, neither is 3^4. But 5^4 is. 5^4=625, the sum of its digits is 11, it's a perfect square (though I slightly forgot what that means). Logic I think!
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lou p lou
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#7
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#7
(Original post by 2776)
I agree with 625, how did u work that out?
I agree with 625, how did u work that out?
*is very tempted to edit post to correct answer to look brainy*
lou xxx
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Juwel
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#8
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#8
(Original post by lou p lou)
yeah, i actually do think it's 625, i just realised that 256 can only go back 3 roots, so there is no 4th root
*is very tempted to edit post to correct answer to look brainy*
lou xxx
yeah, i actually do think it's 625, i just realised that 256 can only go back 3 roots, so there is no 4th root
*is very tempted to edit post to correct answer to look brainy*
lou xxx
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Phil_C
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#9
(Original post by 2776)
I agree with 625, how did u work that out?
I agree with 625, how did u work that out?
Fourth root of 256 is 4 not prime so wrong.
perfect square therefore two whole numbers multiplied together
fourth root prime so whole number form above line is also a perfect square.
ie look for prime no ^4 that is 3 digits, noting 10^4 is 1000 so prime is < 10 ie 2,3,5 or 7
Prime answer is 5, 5^4 is 625
6+2+5 = 13
sqrt 625 is 25
job done answer is 625
incidentally 3^4 is 81 too small
7^4 = 2401 too large!
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Juwel
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#10
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#10
(Original post by Phil_C)
625 is the right answer.
Fourth root of 256 is 4 not prime so wrong.
perfect square therefore two whole numbers multiplied together
fourth root prime so whole number form above line is also a perfect square.
ie look for prime no ^4 that is 3 digits, noting 10^4 is 1000 so prime is < 10 ie 2,3,5 or 7
Prime answer is 5, 5^4 is 625
6+2+5 = 13
sqrt 625 is 25
job done answer is 625
incidentally 3^4 is 81 too small
7^4 = 2401 too large!
625 is the right answer.
Fourth root of 256 is 4 not prime so wrong.
perfect square therefore two whole numbers multiplied together
fourth root prime so whole number form above line is also a perfect square.
ie look for prime no ^4 that is 3 digits, noting 10^4 is 1000 so prime is < 10 ie 2,3,5 or 7
Prime answer is 5, 5^4 is 625
6+2+5 = 13
sqrt 625 is 25
job done answer is 625
incidentally 3^4 is 81 too small
7^4 = 2401 too large!
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Juwel
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#11
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#11
Hang on, you said 7^4 is 4 digits, after saying that 10^4 is the lowest 4 digit number...
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Phil_C
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#12
(Original post by ZJuwelH)
Hang on, you said 7^4 is 4 digits, after saying that 10^4 is the lowest 4 digit number...
Hang on, you said 7^4 is 4 digits, after saying that 10^4 is the lowest 4 digit number...
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Juwel
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#13
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#13
(Original post by Phil_C)
Your quite right, should have been 10^4 = 10000 what has become of my maths, I don't know.
Your quite right, should have been 10^4 = 10000 what has become of my maths, I don't know.
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GH
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#14
Adhsur
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#15
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#15
(Original post by 2776)
REP duley given
REP duley given
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theone
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#16
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#16
Here's a new problem:
Let abcdefghi be a nine digit number such that each of the digits 1,2,3...9 is used once and only once, and the two digit number ab is divisible by 2, 3 digit number abc is divisible by 3, abcd divides 4, etc...
(Note: by abc... I don't mean a.b.c ... but the number with the digits a,b,c...).
Find a possibility for abcdefghi. Is there more than one?
Let abcdefghi be a nine digit number such that each of the digits 1,2,3...9 is used once and only once, and the two digit number ab is divisible by 2, 3 digit number abc is divisible by 3, abcd divides 4, etc...
(Note: by abc... I don't mean a.b.c ... but the number with the digits a,b,c...).
Find a possibility for abcdefghi. Is there more than one?
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Juwel
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#17
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#17
(Original post by theone)
Here's a new problem:
Let abcdefghi be a nine digit number such that each of the digits 1,2,3...9 is used once and only once, and the two digit number ab is divisible by 2, 3 digit number abc is divisible by 3, abcd divides 4, etc...
(Note: by abc... I don't mean a.b.c ... but the number with the digits a,b,c...).
Find a possibility for abcdefghi. Is there more than one?
Here's a new problem:
Let abcdefghi be a nine digit number such that each of the digits 1,2,3...9 is used once and only once, and the two digit number ab is divisible by 2, 3 digit number abc is divisible by 3, abcd divides 4, etc...
(Note: by abc... I don't mean a.b.c ... but the number with the digits a,b,c...).
Find a possibility for abcdefghi. Is there more than one?
Bah forget logic let me just guess...
123456789?
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Unrеgіstered
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#18
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#18
(Original post by Phil_C)
Who am I? I am three digits in length, a perfect square, my fourth root is Prime as is the sum of my digits.
Who am I? I am three digits in length, a perfect square, my fourth root is Prime as is the sum of my digits.
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theone
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#19
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#19
(Original post by ZJuwelH)
b, d, f, h= either 2, 4, 6 or 8.
Bah forget logic let me just guess...
123456789?
b, d, f, h= either 2, 4, 6 or 8.
Bah forget logic let me just guess...
123456789?
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Juwel
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#20
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#20
(Original post by ZJuwelH)
b, d, f, h= either 2, 4, 6 or 8.
Bah forget logic let me just guess...
123456789?
b, d, f, h= either 2, 4, 6 or 8.
Bah forget logic let me just guess...
123456789?
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