Question

Total number of even divisors of $ 2079000 $ which are divisible by $ 15 $ are:
(A) $ 54 $
(B) $ 128 $
(C) $ 108 $
(D) $ 72 $

Answer 1

Hint: In the given question, we are required to find the number of factors of a given number under some specific conditions. So, to find the number of divisors of the number under the specific conditions, we first find the prime factors of the number using the prime factorization method. Then, we keep aside the factors by which each number to be counted has to be divisible and choose from the rest of the factors as the options.

Complete step by step solution:
In the given question, we are required to find the number of even divisors of $ 2079000 $ which are divisible by $ 15 $ .
We know that even numbers are the numbers that are divisible by $ 2 $ .
Now, we have to count the factors of the number $ 2079000 $ that are divisible by $ 15 $ and $ 2 $ .
So, we first express $ 2079000 $ as the product of prime numbers using the prime factorization method.
 $ 2079000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 7 \times 11 $
Expressing in exponents of prime factors, we get,
 $ \Rightarrow 2079000 = {2^3} \times {3^3} \times {5^3} \times 7 \times 11 $
Now, we have to find the factors of $ 2079000 $ which are collectively divisible by $ 30 $ . So, we keep aside $ 30 $ as a factor in the factorization of the number $ 2079000 $ .
So, we get,
 $ \Rightarrow 2079000 = 30 \times \left( {{2^2} \times {3^2} \times {5^2} \times 7 \times 11} \right) $
Now, the remaining exponent of $ 2 $ is $ 2 $ . So, we have a total of three options to select for the power of $ 2 $ : $ 0 $ , $ 1 $ and $ 2 $ .
Similarly, the remaining exponent of $ 3 $ is $ 2 $ . So, we have a total of three options to select for the power of $ 3 $ : $ 0 $ , $ 1 $ and $ 2 $ .
Similarly, the remaining exponent of $ 5 $ is $ 2 $ . So, we have a total of three options to select for the power of $ 5 $ : $ 0 $ , $ 1 $ and $ 2 $ .
Also, the exponent of $ 7 $ is $ 1 $ . So, we have a total of two options to select for the power of $ 7 $ : $ 0 $ and $ 1 $ .
Also, the exponent of $ 11 $ is $ 1 $ . So, we have a total of two options to select for the power of $ 11 $ : $ 0 $ and $ 1 $ .
Now, the total number of options for the combinations of factors when $ 30 $ is kept aside is $ 3 \times 3 \times 3 \times 2 \times 2 $ .
 $ \Rightarrow {3^3} \times {2^2} $
Simplifying the calculations, we get,
 $ \Rightarrow 27 \times 4 $
 $ \Rightarrow 108 $
Hence, the total number of even divisors of $ 2079000 $ which are divisible by $ 15 $ is $ 108 $ .
Therefore, option (C) is correct.
So, the correct answer is “Option C”.

Note: In such questions, we must know the method to keep a factor or number aside in order to calculate the number of combinations possible with the other factors. One must take care of the calculations in order to be sure of the final answer. We should have a strong grip over a variety of problems in permutations and combinations.