If the LCM of a amd 12 is 36 what could be the possible values of a?

Question

If the LCM of A and 12 is 36, what are the possible values of A?

I. 9
II. 18
III. 36

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

KarishmaB · Accepted Answer

Think of what LCM means before going ahead. If I say LCM of two numbers is  36 (= 4*9 = 2^2 * 3^2) , it means that at least one of them must have a  2^2  and at least one of them must have a  3^2  (It is not possible that both numbers have just 3 because then, the LCM would have just 3, not 9)

If one number is  12 (= 2^2 * 3) , the other number A must have  3^2  since 12 doesn't have it.
So minimum value of A will be 9. A can also have a 2 or a  2^2  so other possible values are  18 (=9*2)  and  36 (= 9*2^2) 
Also, A cannot have any other factors since if it did, then the LCM would have to have that factor too.

enigma123 · Answer

If the LCM of a & 12 is 36 what could be the possible values of a?

The LCM of 12 and a contains two 2's. Since the LCM contains each prime factor to the power it appears the MOST, we know that a cannot contain more than two 2's.

LCM of 12 and a contain two 3's. But 12 only contains one 3. The 3^2 factor in the LCM must have come from prime factorization of a. Thus we know that a contains exactly two 3's.

Since a must contain exactly two 3's nd can contain no 2's, one 2 or two 2's a could be

3*3=9
3*3*2=18
3*3*2**2=36.

Thus 9,18 and 36 are three values.

I am struggling to understand the concept.

walker · Answer

Here is what I do in my head (mental tricks):

LCM ( 12 ,36) --> (12/ 12 ,36/ 12 ) --> (1, 3 ) -->  3 * 12  or 1*36 --> 36
LCM ( 12 , 30) --> (12/ 6 ,30/ 6 ) --> (2, 5 ) -->  5 * 12  or 2*30 --> 60
LCM ( 10 , 28) --> (10/ 2 ,28/ 2 ) --> (5, 14 ) -->  14 * 10  or 5*28 --> 140

GCD (12,36) --> (12/ 12 ,36/ 12 ) --> (1,3) -->  12 
GCD (12, 30) --> (12/ 6 ,30/ 6 ) --> (2,5) -->  6 
GCD (10, 28) --> (10/ 2 ,28/ 2 ) --> (5,14) -->  2

LCM and GCD have a nice property: LCM(x,y)*GCD(x,y) = xy

For example, LCM(12,30)*GCD(12,30) = 60*6 = 12 * 30

dimri10 · Answer

i thank both of you, but i did not get walker's approach.can you please sharpen your explenation?

let's say that there are 3 numbers:
440,120 and 80.how can you use your shortcut

walker · Answer

LCM ( 80 , 120, 440) --> (80/ 40 , 120/ 40 , 440/ 40 ) --> (2,  3 ,  11 ) -->  3 * 11 * 80  --> 2640

GCD (80, 120, 440) --> (80/ 40 , 120/ 40 , 440/ 40 ) --> (2, 3, 11) -->  40

40 here is the greatest common divisor.

Spidy001 · Answer

12 = (2^2)*3
a = (2^2 or 2^1 or 2^0)*(3^2)

L.C.M of a & 12 = 36 = (2^2 )*(3^2) = product of Maximum powers of each prime factor

This can be easily understood by comparing 12 with 36 .

12 has one 3 , but L.C.M which is the maximum powers of each prime factor has two 3's

=> a must have two 3's in it = 3^2 ------1

12 has two 2's. L.c.m which is the maximum powers of each prime factor has two 2's as well.

=> a could have zero 2's or one 2 or two 2's------2

clubbing above two statements marked 1 *2 , we have a = (2^2 or 2^1 or 2^0)*(3^2)
 =(4 or 2 or 1)*9
 => a could be 36 or 18 or 9.

Hope its clear now.

anikgupta18 · Answer

Thanks Karishma. Your solution is very clear and in simple way. I didn't understand the complicated solution prepared by Manhattan. Their prime columns technique is good but solution explanation was confusing.

woohoo921 · Answer

KarishmaB

Why can't a potential value for a be just 3?
My reasoning is 3*12=36.

Many thanks!

KarishmaB · Answer

Because LCM is the lowest common multiple. It is the product of the two numbers only when they have no common factors. 
LCM of 3 and 12 will not be 36. It will be 12.

unraveled · Answer

A = x*y*z*....
12 = 2^2*3

LCM(A,12) = 36 = 2^2*3^2

The least common of A and 12 must have highest power factors of each of them that constitute LCM 36
2^2 is with 12 so 3^2 must be with A. 
9 <= A = 9*multiples

However, what multiples we can have. We can have only 2 with highest power of 2. 
Thus, A can be
9*1 = 9
9*2 = 18 
9*2^2 = 36

Any more multiples would not result in LCM of 36.

MagicMeat · Answer

can someone confirm if the answer is E?

Bunuel · Answer

Yes, the OA is E.

If the LCM of A and 12 is 36, what are the possible values of A?

I. 9
II. 18
III. 36

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

We need LCM(A, 12) = 36.

12 = 2^2 * 3
36 = 2^2 * 3^2

So A must contain 3^2 = 9, while introducing no prime factor or power beyond those in 36.

Check the choices:

I. A = 9: LCM(9, 12) = 36. Yes.
II. A = 18: LCM(18, 12) = 36. Yes.
III. A = 36: LCM(36, 12) = 36. Yes.

Answer: E.

If the LCM of a amd 12 is 36 what could be the possible values of a?

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If the LCM of a amd 12 is 36 what could be the possible values of a?

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