Question

Let $N=2^4\times 3^5\times 5^2\times 7^4\times 11$ be resolved as a product of two factors (order immaterial), then

• A. number of cases in which the two factors are coprime is $16$
• B. number of cases in which their H.C.F. is $3$, is $32$
• C. number of cases in which their H.C.F. $5$, is $16$
• D. number of cases in which their H.C.F is odd is $180$

Answer 1

Answer: (A), (D)

number of cases in which their H.C.F is odd is $180$

Number of cases in which two factors are coprime:
For this case, we need to divide the numbers such that two factors do not have any common factor. Let's say the factors are $a$ and $b$ and $a\times b=2^4\times 3^5\times 5^2\times 7^4\times 11$
Here, we need to understand that if $a$ is take $2^{\text{something}}$ as its factor, then that $\text{'something'}$ must be $4$. Because if $a$ takes $2^2$ as factor then the remaining $2^2$ has to be factor of $b$, in this case $a$ and $b$ are not coprime.

Here,
$2^4\to 2\text{choices, either go to}a\text{or}b{3}^5\to 2\text{choices, either go to}a\text{or}b{5}^2\to 2\text{choices, either go to}a\text{or}b{7}^4\to 2\text{choices, either go to}a\text{or}b11\to 2\text{choices, either go to}a\text{or}b$
No. of cases $\to \frac{2^5}{2}=2^4=16\text{we will divide by 2 because order doesn't matter,}a=2^4\times 3^5,b=5^2\times 7^4\times 11\text{is same as}b=2^4\times 3^5,a=5^2\times 7^4\times 11$

Number of cases in which their H.C.F. is $3$:
Let $a$ and $b$ be two parts then $a=3k_1\&b=3k_2$ where $k_1$ and $k_2$ are coprime
Remaining ways $\to \frac{2^5}{2}=16$

Number of cases in which their H.C.F. is $5$:
no. of cases$\to \frac{2^4}{2}=8$

Number of cases in which their H.C.F. is odd:
no. of cases $=\frac{2\times 6\times 3\times 5\times 2}{2}=180$

as $\begin{array}{c}{2}^4\text{has}2\text{ways}\\ 3^5\text{has}6\text{ways}\\ 5^2\text{has}3\text{ways}\\ 7^4\text{has}5\text{ways}\\ {11}^1\text{has}2\text{ways}\end{array}\}\text{of dividing into}2\text{factors}$