Question

The number of even divisors of the number \[N=12600={{2}^{3}}{{3}^{2}}{{5}^{2}}7\] is
(a) 72
(b) 54
(c) 18
(d) none of these

Answer 1

Hint: We will first assume the power of 2 as a, power of 3 as b, power of 5 as c and power of 7 as d. Then we will define the intervals in which a, b, c and lies. As even divisors have been asked we will remove 0 as the power of a because 2 will always be there.

Complete step-by-step answer:
It is mentioned in the question that \[N=12600={{2}^{3}}{{3}^{2}}{{5}^{2}}7\].
So let the power of 2 be a and it lies between 0 and 3 that is \[0\le a\le 3\].
And now let the power of 3 be b and it lies between 0 and 2 that is \[0\le b\le 2\].
Also assuming the power of 5 to be c and it lies between 0 and 2 that is \[0\le c\le 2\].
Finally assuming the power of 7 to be d and it lies between 0 and 1 that is \[0\le d\le 1\].
So in the question number of even divisors has been asked which means 2 will always be there and hence removing 0 from a, we get the elements of a as 1, 2, 3.
Hence the number of elements in a is \[n(a)=3......(1)\]
Now elements in b is 0, 1, 2 and hence the number of elements in b is \[n(b)=3......(2)\]
Now elements in c is 0, 1, 2 and hence the number of elements in c is \[n(c)=3......(3)\]
Now elements in d is 0, 1 and hence the number of elements in d is \[n(d)=2......(4)\]
So from equation (1), equation (2), equation (3) and equation (4) we get,
Total number of even divisors \[=n(a)\times n(b)\times n(c)\times n(d).......(5)\]
Hence substituting the values in equation (5) we get,
\[\Rightarrow 3\times 3\times 3\times 2=54\]
Hence the number of even divisors is 54. So the correct answer is option (b).

Note: We in a hurry can make a mistake in counting the elements of a as we might take 0 also and then the number of elements of a would have been 4 and we would have got 72 as the answer. Because of this reason we need to be careful at what question is asking.