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Functions question

2023 · 30 Jan · Shift 2 · Q36

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Functions
2023 · 30 Jan · Shift 2 · Q36

Functions question

2023 · 30 Jan · Shift 2 · Q36

JEE MainMathematicsFunctionsNumerical+4 / −1

Let

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

. Then the number of possible functions

f: A \rightarrow 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

\underline{\hspace{2cm}}

.

Numerical answer

View written solutionFree

Correct answer: 432

We need to count functions $f:A\to A$ , where $A=\{1,2,3,5,8,9\},$ such that $f(mn)=f(m)f(n)$ for every $m,n\in A$ with $mn\in A$ .

So first, identify all products of elements of $A$ that still lie in $A$ .

Check possible multiplicative relations inside $A$ :

$1\cdot a=a\in A$ for every $a\in A$ .
$2\cdot 2=4\notin A$ .
$2\cdot 3=6\notin A$ .
$2\cdot 5=10\notin A$ .
$2\cdot 8=16\notin A$ .
$2\cdot 9=18\notin A$ .
$3\cdot 3=9\in A$ .
$5\cdot 5=25\notin A$ .
$2\cdot 4$ not relevant since $4\notin A$ .
$2\cdot 2\cdot 2=8$ , but the condition is only for pairs; however $8=2\cdot 4$ does not help because $4\notin A$ .

Also, $8=2\cdot 4 \quad (4\notin A), \qquad 9=3\cdot 3.$ So the only nontrivial pair relation is $f(9)=f(3)^2.$

From multiplication by $1$ , we also get constraints.

Use the condition with $m=1$ and any $n\in A$ : $f(1\cdot n)=f(1)f(n).$ Since $1\cdot n=n$ , this gives $f(n)=f(1)f(n) \quad \text{for all } n\in A.$ Because $f(n)\in A$ and no element of $A$ is $0$ , we can divide by $f(n)$ to get $f(1)=1.$ So $f(1)$ is forced.
Now use the only other nontrivial relation: $f(9)=f(3)^2.$ Since $f(9)\in A$ , we need $f(3)^2\in A$ . Let us test possible values of $f(3)\in A$ :

If $f(3)=1$ , then $f(9)=1^2=1\in A$ .
If $f(3)=2$ , then $f(9)=4\notin A$ .
If $f(3)=3$ , then $f(9)=9\in A$ .
If $f(3)=5$ , then $f(9)=25\notin A$ .
If $f(3)=8$ , then $f(9)=64\notin A$ .
If $f(3)=9$ , then $f(9)=81\notin A$ .

Hence only two possibilities: $f(3)=1 \Rightarrow f(9)=1,$ $f(3)=3 \Rightarrow f(9)=9.$ So there are exactly $2$ choices for the pair $(f(3),f(9))$ .

What about $f(2), f(5), f(8)$ ? There are no multiplicative relations in $A$ involving these elements that produce another element of $A$ (except multiplication by $1$ , which only forced $f(1)=1$ ). So each of $f(2), f(5), f(8)$ can be chosen freely from $A$ .

Thus:

$f(2)$ : $6$ choices
$f(5)$ : $6$ choices
$f(8)$ : $6$ choices
$(f(3),f(9))$ : $2$ choices
$f(1)$ : $1$ forced choice

Total number of functions: $1\cdot 6\cdot 6\cdot 6\cdot 2=432.$

Therefore, the required number of functions is $\boxed{432}.$

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