# CAT 2020 Set-1 | Question: 51

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How many $3-$digit numbers are there, for which the product of their digits is more than $2$ but less than $7$?

Given that, the product of $3 \text{-digit}$ numbers is more than $2$ but less than $7.$

Let the $3\text{-digit}$ number be $xyz.$ Then, $2< x \times y \times z < 7$

$\Rightarrow x \times y \times z = 3\; \text{(or)}\; 4\;\text{ (or)}\; 5 \;\text{(or)}\; 6$

$\textbf{Case 1:}$ When  $x \times y \times z = 3$

• $\underline{x} \quad \underline{y} \quad \underline{z}$
• $1 \quad 1 \quad 3$
• $1 \quad 3 \quad 1$
• $3 \quad 1 \quad 1$

Three possibilities.

$\textbf{Case 2:}$ When  $x \times y \times z = 4$

• $\underline{x} \quad \underline{y} \quad \underline{z}$
• $1 \quad 1 \quad 4$
• $1 \quad 4 \quad 1$
• $4 \quad 1 \quad 1$
• $1 \quad 2 \quad 2$
• $2 \quad 1 \quad 2$
• $2 \quad 2 \quad 1$

Six possibilities.

$\textbf{Case 3:}$ When  $x \times y \times z = 5$

• $\underline{x} \quad \underline{y} \quad \underline{z}$
• $1 \quad 1 \quad 5$
• $1 \quad 5 \quad 1$
• $5 \quad 1 \quad 1$

Three possibilities.

$\textbf{Case 4:}$ When  $x \times y \times z = 6$

• $\underline{x} \quad \underline{y} \quad \underline{z}$
• $1 \quad 2 \quad 3$
• $1 \quad 3 \quad 2$
• $2 \quad 3 \quad 1$
• $2 \quad 1 \quad 3$
• $3 \quad 1 \quad 2$
• $3 \quad 2 \quad 1$
• $1 \quad 1 \quad 6$
• $1 \quad 6 \quad 1$
• $6 \quad 1 \quad 1$

Nine possibilities.

Total numbers $= 3 + 6 + 3 + 9 = 21.$

$\therefore$ There are $21$ numbers, whose product of their digits is more than $2$ but less than $7.$

Correct Answer$: 21$

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