338c=r^3

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]

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!!!!

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

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)

is this some sort of a trick/formula ???

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]

bhumika.k.shah wrote:is this some sort of a trick/formula ???

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

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]

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]

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?

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

Aite! so u got ur mistake!

now explain from the start!

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

bhumika.k.shah wrote:Aite! so u got ur mistake!

now explain from the start!

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

ajith wrote: Well, I think I have given it my best shot

i agree.