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338c=r^3

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by ajith » Sat Jan 30, 2010 1:06 am
bhumika.k.shah wrote:If 338c = r^3 and c is the smallest positive integer such that r is an integer, then r must be

(A) 2
(B) 13
(C) 26
(D) 52
(E) 104
338 = 2*169 =2*13*13

now to make 338 a cube it should be multiplied with at least 2^2*13
[spoiler]
52 is the answer[/spoiler]
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by bhumika.k.shah » Sat Jan 30, 2010 1:07 am
338 = 2*169 =2*13*13 - GOT IT!

now to make 338 a cube it should be multiplied with at least 2^2*13 - why?????

52 is the answer[/quote] -NOPES!!!!
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by thephoenix » Sat Jan 30, 2010 1:12 am
ajith wrote:
bhumika.k.shah wrote:If 338c = r^3 and c is the smallest positive integer such that r is an integer, then r must be

(A) 2
(B) 13
(C) 26
(D) 52
(E) 104
338 = 2*169 =2*13*13

now to make 338 a cube it should be multiplied with at least 2^2*13
[spoiler]
52 is the answer[/spoiler]
i think c=52 and r=26

imo C sud be the ans
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by ajith » Sat Jan 30, 2010 1:14 am
bhumika.k.shah wrote:338 = 2*169 =2*13*13 - GOT IT!

now to make 338 a cube it should be multiplied with at least 2^2*13 - why?????

52 is the answer
-NOPES!!!![/quote]


Because r^3 should be represented in a^3*b^3

manner and by multiplying 338 with 52 it makes it 2^3*13^3

which has an integer cube root of 2*13 (hence r =26 and c =52)

r= 26 (sorry for the mistake in the previous post, I thought they were asking for the value of c)
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by bhumika.k.shah » Sat Jan 30, 2010 1:19 am
is this some sort of a trick/formula ??? :-P
ajith wrote:
bhumika.k.shah wrote:338 = 2*169 =2*13*13 - GOT IT!

now to make 338 a cube it should be multiplied with at least 2^2*13 - why?????

52 is the answer
-NOPES!!!!

Because r^3 should be represented in a^3*b^3

manner and by multiplying 338 with 52 it makes it 2^3*13^3

which has an integer cube root of 2*13 (hence r =26 and c =52)

r= 26 (sorry for the mistake in the previous post, I thought they were asking for the value of c)[/quote]
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by ajith » Sat Jan 30, 2010 1:25 am
bhumika.k.shah wrote:is this some sort of a trick/formula ??? :-P
Probably a trick!

Any number which has integer cuberoot can be shown to be, r^3

Find the current composition of the number say x = a^x*b^y*c^z..(format- a,b,c ....being prime)

to make it a cube by multiplying with another number u need to multiply it to make x,y,z...(all of them) a multiple of 3
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by sanju09 » Sat Jan 30, 2010 3:05 am
bhumika.k.shah wrote:If 338c = r^3 and c is the smallest positive integer such that r is an integer, then r must be

(A) 2
(B) 13
(C) 26
(D) 52
(E) 104
ajith is right at 338*c =2*13*13*c = r*r*r

It at least need two 2's and one 13 for 338*c to become a perfect cube of any integer; and the supply makes 338*c look like 2*2*2*13*13*13 = 2^3*13^3 = (2*13) ^3 = (26) ^3 = (r) ^3

Hence, r = [spoiler]26[/spoiler].

[spoiler]C[/spoiler]
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by bhumika.k.shah » Sat Jan 30, 2010 8:08 pm
Thanks to all!
sanju09 wrote:
bhumika.k.shah wrote:If 338c = r^3 and c is the smallest positive integer such that r is an integer, then r must be

(A) 2
(B) 13
(C) 26
(D) 52
(E) 104
ajith is right at 338*c =2*13*13*c = r*r*r

It at least need two 2's and one 13 for 338*c to become a perfect cube of any integer; and the supply makes 338*c look like 2*2*2*13*13*13 = 2^3*13^3 = (2*13) ^3 = (26) ^3 = (r) ^3

Hence, r = [spoiler]26[/spoiler].

[spoiler]C[/spoiler]
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by sanju09 » Sun Jan 31, 2010 10:18 pm
What's all that, ajith?
Any number which has integer cuberoot can be shown to be, r^3


Where r is...?
Find the current composition of the number say x = a^x*b^y*c^z..(format- a,b,c ....being prime)
Do you mean the same x? What are x, y, and z here?
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by ajith » Sun Jan 31, 2010 10:31 pm
sanju09 wrote:What's all that, ajith?
Any number which has integer cuberoot can be shown to be, r^3


Where r is...?

r is an integer
Find the current composition of the number say x = a^x*b^y*c^z..(format- a,b,c ....being prime)
Do you mean the same x? What are x, y, and z here?
Oh no, not the same x... That is a mistake

m = a^x*b^y*c^z..(format- a,b,c ....being prime)

x, y, z are integers and again if the number m, has a cube root all of x,y,z.. will be a multiple of 3
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by bhumika.k.shah » Mon Feb 01, 2010 3:38 am
Aite! so u got ur mistake!

now explain from the start! :-P
ajith wrote:
sanju09 wrote:What's all that, ajith?
Any number which has integer cuberoot can be shown to be, r^3


Where r is...?

r is an integer
Find the current composition of the number say x = a^x*b^y*c^z..(format- a,b,c ....being prime)
Do you mean the same x? What are x, y, and z here?
Oh no, not the same x... That is a mistake

m = a^x*b^y*c^z..(format- a,b,c ....being prime)

x, y, z are integers and again if the number m, has a cube root all of x,y,z.. will be a multiple of 3
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by ajith » Mon Feb 01, 2010 3:48 am
bhumika.k.shah wrote:Aite! so u got ur mistake!

now explain from the start! :-P
Well I was trying explain a method by which we can make an integer a cube

Step 1. Factorize the given number to prime factors in the format a^x*b^y*c^z....


in this case the number is 338 it can be represented as = 2^1*13^2

a =2 b = 13 x =1 y =2

Step 2:

For a number to become a cube x,y,.... all should be a multiple of 3

the closest multiple of 3 near to x,y is 3

Step 3:

Now to make it a cube we have to multiply 338 with 2^(3-x)*13^(3-y)

ie 2^2*13^1 = 4*13 =52

Well, I think I have given it my best shot
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by bhumika.k.shah » Mon Feb 01, 2010 3:53 am
ajith wrote: Well, I think I have given it my best shot
i agree. :-)
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