What is the value of integer x?

Excellent Question.
Here is my approach
From 1 => LCM of x and 45 = 225 => x can have 3 values => 5^2
5^2*3
5^2*3^2
=> not sufficient
From 1 => LCM of x and 20 = 300 => x can have 3 values => 5^2*3
5^2*3*2
5^2*3*2^2 => not sufficient
Combining them we can clearly say that x=5^2*3

SMASH C

STONE COLD STEVE AUSTIN

Hi,

Thanks for pointing it out. I have corrected the Typo in earlier response. Same is highlighted.

The concept being tested here is Prime factorization. Also, it tests how the LCM of two numbers is derived. Let me attempt to solve this long hand.

S1:
Prime Factors of 45 (One of the two numbers) = \(3^2\),\(5^1\)
Prime factors of 225 (LCM of the two numbers) = \(3^2\), \(5^2\)
If the LCM of X and 45 is 225, what are different possible prime factors of X?
* X HAS to have \(5^2\) as one of its factors
* X COULD have \(3^0\),\(3^1\) or \(3^2\) as its other factors
* Possible values of X are -
-> \(5^2*3^0 = 25\)
-> \(5^2*3^1 = 75\)
-> \(5^2*3^2 = 225\)
Three Answers for X and hence Insufficient.

S2:
Prime Factors of 20 (One of the two numbers) = \(2^2\),\(5^1\)
Prime factors of 300 (LCM of the two numbers) = \(2^2\),\(3^1\), \(5^2\)
If the LCM of X and 20 is 300, what are different possible prime factors of X?
* X HAS to have \(5^2\) as one of its factors
* X has to have \(3^1\) as one of its factors
* X COULD have \(2^0\),\(2^1\) or \(2^2\) as its other factors
* Possible values of X are -
-> \(5^2*3^1*2^0 = 75\)
-> \(5^2*3^1*2^1 = 150\)
-> \(5^2*3^1*2^2 = 300\)
Three Answers for X and hence Insufficient.

S1+S2

The only common values between S1 and S2 = \(5^2*3^1*2^0 = 75\)

Hence the Answer choice C

I hope this helps.

PS:

Recommendation: Please read the MGMAT Number properties guide on prime factors. This approach is discussed thoroughly there. It will go a long way in clearing your concepts on Prime numbers - among others.

HI,

This is how it is

S1: Possible values = 25 or 75 or 225 : insuff
S2: Possible Values = 75 or 150 or 300 : insuff
s1+ S2 : Only one common value = 75 : Suff
Answer choice : C

We of course don't need to know the value is 75. Just knowing there is a unique common value between S1 and S2 is enough.

Hope this helps.

Hello triple4!

This might be a long answer. Please bear with me. Intention is to clarify and not confuse

To understand this we need to be clear about the 'Prime column' approach for finding the LCM of two numbers. This is very nicely explained in MGMAT number guide books.

• Firstly, lets see what are the factors of one of the numbers (lets call this the other number) - \(20 = 2^2, 3^0, 5^1\)
  
• Secondly, lets see what are the factors of the LCM = \(300 = 2^2,3^1, 5^2\)

The thing to know here is that the LCM of two numbers are formed by taking the Highest Power of the Prime factors of the numbers involved. We have three prime factors to consider here 2,3 and 5.

• The LCM has two fives (\(5^2\)) and the other number has only one five (\(5^1\)). The highest power of five is two. This means the LCM got the two five's from X. Meaning one of the factors of X = \(5^2\)
• The LCM has one three (\(3^1\)) and the other number has Zero threes. The highest power of three is one. This means the LCM got the three from X. Meaning one of the factors of X = \(3^1\)
• The LCM has two two's (\(2^2\)) and the other number also has two two's. This means LCM COULD have got both the two's from X or the other number. There are three ways in which this can happen. a) If X has no two's (\(2^0\)) b) X has one two (\(2^1\)) or c) X also has two two's (\(2^2\)).

Combining all the above can say X could be = \(5^2*3^1*2^0\) or \(5^2*3^1*2^1\) or \(5^2*3^1*2^2\)

I hope this made sense. If no, please PM me and I will try to explain. The entire thing is well covered in "Extra Divisibility and Primes" chapter of MGMAT. This is explained under “Finding GCF and LCM Using Prime Columns"

Hope this helped!

Consider this: LCM may not be the product of 2 numbers. If they have some common factors, LCM will count them only once.
Eg LCM of 10 and 5 is 10, not 50.
Also multiplying the two diff values of x obtained will not give a value of x.

LCM of x and 45 is 225.
45 = 3*3*5
225 = 3*3*5*5

So we can say for sure that x has 5*5 = 25 as a factor. It may or may not have one or two 3s as factors too but nothing else.

LCM of x and 20 is 300.
20 = 2*2*5
300 = 2*2*3*5*5

So we can say for sure that x has 3*5*5 = 75 as a factor. It may or may not have one or two 2s as factors too but nothing else.

Using both, we see that x must be 75.

I understand how you get to 75, but how can you be sure x doesn't also have prime factors that are in neither 225 or 300?

To rephrase, I can see how it could be 75, but why does it have to be 75 and not something like 75*11?

In S1: I don't see 225 as 125 as LCM of 25 and 225.

Hi,

I went through this question several times and still can't understand a part of it. Ive also tried to look it up in the MGMAT book.

When you said:

If the LCM of X and 20 is 300, what are different possible prime factors of X?
* X HAS to have \(5^2\) as one of its factors
* X has to have \(3^1\) as one of its factors
* X COULD have \(2^0\),\(2^1\) or \(2^2\) as its other factors

I don't understand why X HAS to have 5^2 and why X COULD have 2^0, 2^1 or 2^2.

hi

45x = 225
x = 5

20x = 300
x = 15

x= 15*5 = 75

cheers, if this helps ..

hi mam
thank you.

Yes, I know this way of finding LCM, but
I gave it a try differently and somehow it worked..

Anyway, mam, you are so great ...

We need to determine the value of x.

Statement One Alone:

The LCM of x and 45 is 225.

We can start by breaking 45 and 225 into prime factors.

45 = 3^2 x 5^1

225 = 3^2 x 5^2

In other words, 3^2 x 5^1 and some number x have an LCM of 3^2 x 5^2.

Thus, x could be the following:

5^2 = 25

5^2 x 3^1 = 75

5^2 x 3^2 = 225

Statement one is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

The LCM of x and 20 is 300.

We can start by breaking 20 and 300 into prime factors.

20 = 2^2 x 5^1

300 = 2^2 x 3^1 x 5^2

In other words, 2^2 x 5^1 and some number x have an LCM of 2^2 x 3^1 x 5^2.

Thus, x could be the following:

3^1 x 5^2 = 75

3^1 x 5^2 x 2^1 = 150

3^1 x 5^2 x 2^2 = 300

Statement two is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

Using statements one and two we see that the only value of x that fulfills both statements is 75.

Answer: C

Statement 1- 45= 3*3*5; 225= 3*3*5*5
therefore, x could be : 3*3*5*5 or 3*5*5 or 5*5 we don't know (not sufficient)
Statement 2- 20= 2*2*5; 300= 5*5*2*2*3 (again too many possibilities so not sufficient)
combining 1 and 2, x must have two 5's and one 3 therefore, x=75= 3*5*5
Answer is C

What is the value of integer x?

1. The LCM of x and 45 is 225
2. The LCM of x and 20 is 300.

1) 225 = 5 * 45 = 5 * 5* 9 = 5 * 5 * 3* 3
x can be 25, 75 or 225.

2) 300 = 3 * 10* 10 = 3 * 2* 2 * 5* 5. x can be 75 ,150 or 300.

Together, x is 75.

C is the answer.

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