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. $\frac{11}{180}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{29}{180}$

If a number is divisible by $4$ and $5$, then it is divisible by $20$(since $4$ and $5$ are relatively prime).

A number is divisible by $20$ if it ends in $20,40,60,80$ or $00$.

There exist two cases. Numbers ending in $20,40,60,80$, and numbers ending in $00$.

I) Number of three-digit numbers divisible by $20$ ending in $20,40,60,80=4\times 9=36$

Possible permutations of each of these three-digit number, including the number itself $=4$ (e.g. $520\to 502,205,250$)

Therefore, numbers in this case = $36\times 4=144$

However, when both non-zero digits are even, the permutations are repeated. (e.g. $820\to 802,208,280$ and $280\to 802,208,820$)

Such numbers $=\frac{4\times 4}{2}=8$

Subtract these cases.

$\Rightarrow$ total numbers in this case $=144-8=136$

II) For those numbers ending in $00$, no other permutations are possible. Hence, numbers in this case $=9$

Total cases $=136+9=145$

Total $3$-digit numbers $=9\times 10\times 10=900$

Probability $=\frac{145}{900}=\frac{29}{180}$