Question

If a $3$-digit number is randomly chosen, what is the probability that either the number itself or some permutation of the number (which is a $3$-digit number) is divisible by $4$ and $5$?

• A. $\frac{1}{45}$
• B. $\frac{29}{180}$
• C. $\frac{11}{60}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{11}{60}$

The $3$ digit numbers divisible by both $4$ and $5$ must have last two digits as $00,20,40,60$ or $80$.

The numbers ending in ${}^{\prime }{00}^{\prime }$ are $\{100,200,300,\dots ,900\}$. No permutation of above numbers are possible.
Hence, number of numbers ending with ${}^{\prime }{00}^{\prime }$ are $9$.

If one digit of the number is repeated, e.g. $220$, it cann be permutated in $2$ ways i.e. $220,202$
Hence, for numbers $\{220,440,660,880\}$, we will have $8$ possible numbers.

If none of the digit is repeated, e.g. $420$, it can be permutated in $4$ ways. $(420,402,204,240)$

For the numbers, the number of numbers possible will be $8\times 4\times 4=128$

Therefore, total number of numbers are $145$

Therefore, required probability $=\frac{145}{900}=\frac{29}{180}$