smallest possible value of n

[wawan_andrean]

If both 5^2 and 3^3 are factors of n x 2^5 x 6^2 x 7^3, what is the

smallest possible value of n?

a. 25

b. 27

c. 45

d. 75

e. 125

[magnetronclassic]

Simplifying the expression we get

n*[2][/5]**[3][/2]*[7][/3]

=n*[3][/2]*k

As [5][/2] and are factors the smallest possible value of n=[5][/2]*3=75

[rajatmeh]

Yes (D).

[wawan_andrean]

still stumped..pls explain..

[Project700Plus]

Let me try:

 

So 5^2 and 3^3 are factors of n*2^5*6^2*7^3

 

Rewrite n*2^5*6^2*7^3 as n*2^5*(2*3)^2*7^3 = n*2^5*2^2*3^2*7^3

 

Since 3^2 is already a factor of n*2^5*6^2*7^3, you only need 3*5^2 for n so that 5^2 and 3^3 are factors.

 

Thus smallest value of n=5^2*3=75.

 

Hope this helps.

[genius_in_the_gene]

n*2^5*6*2*7*3

 

Break this into prime factors

n*2^5*2*2^2*3^2*7*3

 

Now since 5^2 and 3^3 are factors of this, n should be a multiple of 5^2( as there are no factors of 5 in the original expression). Also since we are expecting atleast 3 factors of 3 and we have just 2, so n should be a multiple of 3.

 

ANS 25*3=75

[wawan_andrean]

thx all, i got it...

[hardikrs]

sorry to be redundant, but can somebody explain it one more time. i got to the part where we bork it down to prime factors

 

but did not follow this --

 

"Now since 5^2 and 3^3 are factors of this, n should be a multiple of 5^2( as there are no factors of 5 in the original expression). Also since we are expecting atleast 3 factors of 3 and we have just 2, so n should be a multiple of 3."

 

thanks,

[subodhg]

Hardikrs,

 

For the expression n*2^5*2*2^2*3^2*7*3 (which is the original expression expressed in prime factors)

to be divisible by 5^2 and 3^3,

n should be divisible by 5^2 and 3. ( 3^2 ) is already present in the above expression.

 

so n has to be 5^2*3

 

HTH.

[hardikrs]

thanks subodhg.