The function f(x) is defined for all positive integers x as

Answer :-
a=2 and b=3 satisfy both the given conditions. f(b)*g(a)=0
Even number of factors of (3)*Number of odd factors of (2)=0*1=0 &
f(a)=f(2)= Number of even factors of 2=1

Least Common Multiple of 2 & 3 from given options=12
Hence Ans=A

The LCM of 2 and 3 is 6, not 12.

Answer D
a=2, b=15 satisfy both equations
derived these values by factoring answer options
12=4*3
16=4*4
20=4*5
30=2*15
36=2*2*3*3

f(x)= even factors
g(x) = odd factors

now f(a)=1 ; 2 is the only such number which satisfy this condition. hence a=2

g(a)=g(2)=1 as 2 has only one odd factor which is number 1 itself.
this means f(b) must be zero. i.e. b must be odd.

now the highest power of 2 in a and b is 1, therefore L.C.M must contain 1 as the highest power of 2 i.e. 2^1.

now coming to the options only option D satisfy this condition hence answer must be D

f(x)=number of even factors of x,henceforth,f(b)=number of even factors of b and f(a)=number of even factors of a

g(x)=number of odd factors of x,henceforth,g(a)=number of odd factors of a,


f(b).g(a)=0
number of even factors of b .number of odd factors of a=0
it means either the number of even factors of b or the number of odd factors of a is 0
but all the integers at least contain one odd integer 1 as factors,so it indicates number of even factors of b is 0,which means b doesnot contain
any 2 in its prime factorization

f(a)=1
it means a contains only one 2 in its prime factorization (for example a could be 2,6,10,14,18…….)

As per the definition of LCM,it consists of highest power of all the factors of two or more integers
therefore,the LCM will contain only single power of 2(2^1 )

among all the answer choice only D contains single power of 2,therefore,D is the correct answer

SOLUTION

The function f(x) is defined for all positive integers x as the number of even factors of x and the function g(x) is defined for all positive integers x as the number of odd factors of x. For positive integers a and b if f(b)*g(a) = 0 and f(a) = 1, which of the following could be the least common multiple of a and b?

A. 12
B. 16
C. 20
D. 30
E. 36

\(f(b)*g(a) = 0\): any positive integer has at least one odd factor: 1. Thus, g(a) cannot be 0, which implies that f(b) = 0. This on the other hand means that b is an odd integer (odd integers does not have even factors).

\(f(a) = 1\): \(a\) has 1 even factor. This means that \(a\) must be 2 (the only positive integer which has only 1 even factor is 2).

The least common multiple of \(a=2\) and \(b=odd\) is \(2*odd\). Only option D can be represented this way: \(30=2*15\).

Answer: D.

Kudos points given to correct solutions above.

Try NEW divisibility DS question.

Need not be, no?

Even 6 has only one even factor, so does 10, 14, 18?

6 has two even factors 2 and 6.

Awww!

I can't believe I thought otherwise!

Thanks

Given:

f(a) = 1 => a has one even factor and that is 2 and it is must be of form 2 * m (where m is any odd integer)
f(b) * g(a) = 0 => f(b) = 0 OR g(a) = 0
=> but g(a) = 0 means no odd factor, this cannot be so, as every integer 1 as odd factor, so g(x) >= 1
=> so f(b) = 0 => b is odd number

so a : 2 * odd
b : odd

LCM (a, b) must have only one 2. Only D has single 2 as its factor

Answer (D)

I did not take into account that 1 is a factor of every number, so this took me too long (4 mins) but here's the solution:

#EVEN factor a = 1 ... from this we know a = 2 since 2 is the only number with 1 even factor.

#EVEN factor b * #ODD factor a = 0 ... we know that either b has no EVEN factors or a has no ODD factors (Bunuel's solution eliminates the issue I had here with multiple cases...)

Since we know a=2 then #ODD factor a = 1, so the number #EVEN in b must be 0 so that the above equation works.

This means a=2 and b could = 3. Not a solution based on answer choices (LCM would be 6).
Since b can only have odd factors the next one is 3*5 = 15, so 2*15 = 30, which is AC D.

If f(b) = 0 ; the number of even factors of b = 0; b = odd
If g(a) = 0 ; the number of odd factors of a = 0; 1 is odd; not feasible
b = odd

f(a) = 1; the number of even factors of a = 1; a=2o where o if an odd number;

LCM(a,b) = 2o form

IMO D

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