This discussion is closed.
Phil_C
Badges: 7
Rep:
?
#1
Report Thread starter 17 years ago
#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.
0
kyuuki
Badges: 0
Rep:
?
#2
Report 17 years ago
#2
some kind of robotic squirrel?
0
Juwel
Badges: 18
Rep:
?
#3
Report 17 years ago
#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.
625?
0
lou p lou
Badges: 2
Rep:
?
#4
Report 17 years ago
#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.
256?

lou xx
0
GH
Badges: 13
Rep:
?
#5
Report 17 years ago
#5
I agree with 625, how did u work that out?
0
Juwel
Badges: 18
Rep:
?
#6
Report 17 years ago
#6
(Original post by 2776)
I agree with 625, how did u work that out?
Worked on the primes bit.
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!
0
lou p lou
Badges: 2
Rep:
?
#7
Report 17 years ago
#7
(Original post by 2776)
I agree with 625, how did u work that out?
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
0
Juwel
Badges: 18
Rep:
?
#8
Report 17 years ago
#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
<huge grin> Genius!!!
0
Phil_C
Badges: 7
Rep:
?
#9
Report Thread starter 17 years ago
#9
(Original post by 2776)
I agree with 625, how did u work that out?
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!
0
Juwel
Badges: 18
Rep:
?
#10
Report 17 years ago
#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!
Je suis genius!!! Can I have rep for this???
0
Juwel
Badges: 18
Rep:
?
#11
Report 17 years ago
#11
Hang on, you said 7^4 is 4 digits, after saying that 10^4 is the lowest 4 digit number...
0
Phil_C
Badges: 7
Rep:
?
#12
Report Thread starter 17 years ago
#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...
Your quite right, should have been 10^4 = 10000 what has become of my maths, I don't know.
0
Juwel
Badges: 18
Rep:
?
#13
Report 17 years ago
#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.
Imagine the negative effects on a youngster reading that... thank god I clarified it, I think I should be rewarded and compensated with rep for my troubles...
0
GH
Badges: 13
Rep:
?
#14
Report 17 years ago
#14
REP duley given
0
Adhsur
Badges: 15
Rep:
?
#15
Report 17 years ago
#15
(Original post by 2776)
REP duley given
Not that your rep will actually make a difference though...sorry to point out. Ah, but it's the thought that counts!
0
theone
Badges: 0
Rep:
?
#16
Report 17 years ago
#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?
0
Juwel
Badges: 18
Rep:
?
#17
Report 17 years ago
#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?
b, d, f, h= either 2, 4, 6 or 8.

Bah forget logic let me just guess...

123456789?
0
Unrеgіstered
Badges: 0
#18
Report 17 years ago
#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.
a freak
0
theone
Badges: 0
Rep:
?
#19
Report 17 years ago
#19
(Original post by ZJuwelH)
b, d, f, h= either 2, 4, 6 or 8.

Bah forget logic let me just guess...

123456789?
Nope, 1234 is not divisible by 4
0
Juwel
Badges: 18
Rep:
?
#20
Report 17 years ago
#20
(Original post by ZJuwelH)
b, d, f, h= either 2, 4, 6 or 8.

Bah forget logic let me just guess...

123456789?
Oops wrong, abcdefg does not divide by 7!
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Following the government's announcement, do you think you will be awarded a fair grade this year?

Yes (480)
51.84%
No (446)
48.16%

Watched Threads

View All