Let $\mathrm{A}=\{1,2,3,5,8,9\}$. Then the number of possible functions $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{A}$ such that $\mathrm{f}(\mathrm{m} \cdot \mathrm{n})=\mathrm{f}(\mathrm{m}) \cdot \mathrm{f}(\mathrm{n})$ for every $\mathrm{m}, \mathrm{n} \in \mathrm{A}$ with $\mathrm{m} \cdot \mathrm{n} \in \mathrm{A}$ is equal to \_\_\_\_ .

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Sets Relations and Functions

Let

A = {1, 2, 3, 5, 8, 9}

. Then the number of possible functions

f : A \to A

such that

f (m \cdot n) = f (m) \cdot f (n)

for every

m, n \in A

with

m \cdot n \in A

is equal to ____ .

Answer: 432

From

f (1) = f (1 \cdot 1) = f (1)^{2}

and

f (1) \in A

f (1) = 1

. The only product of two elements of

A

that stays in

A

(apart from multiplications by

1

) is

3 \cdot 3 = 9

, so

f (9) = f (3)^{2}

; for

f (9) \in A

this needs

f (3) \in {1, 3}

, and then

f (9)

is determined. The elements

2, 5, 8

carry no such relation, so

f (2), f (5), f (8)

are free with

6

choices each. The count is

1 \times 6 \times 2 \times 6 \times 6 \times 1 = 432

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Integer answer