KVPY 2016 SA Question 22

The number of 3-digit integers is 100 to 999 hence $N=900$ , the number of ways to choose a 3-digit integers.

The required integer is one with a permutation divisible by 4 and 5 or divisible by 20. Consider a 3-digit number $\overline{ab0}$ . Then $\dfrac {\overline{ab0}}{20} = \dfrac {\overline{ab}}{2}$ . We note that $\overline{ab}$ is divisible by 2 only if $b$ is even. Therefore the permutation must end with 00, 20, 40, 60 or 80. There are four cases to consider.

• Case 00: Since 0 can not be the hundred digit, then there are only 9 cases, 100, 200, 300, ..., 900 $\implies N_{00} = 9$ .
• Case 120 Odd-even-zero: Since 0 can only be the ten or unit digit, there are 4 cases 102, 120, 201 and 210. Since there are 5 odd numbers and 4 even numbers $\implies N_{120} = 5 \times 4 \times 4 = 80$ .
• Case 220 Same-even-zero: For same even numbers, there are only 2 cases, 202 and 220. Since there are 4 even numbers $\implies N_{220} = 4 \times 2 = 8$ .
• Case 420 Even-even-zero: There are 4 cases for a pair of even numbers, 204, 240, 402 and 420. Since there are ${4 \choose 2}$ ways to choose the pair of even numbers, then $\implies N_{420} = {4 \choose 2} \times 4 = 6 \times 4 = 24$ .

Then the probability $Pr = \dfrac {N_{00}+N_{120}+N_{220}+N_{420}}N = \dfrac {9+80+8+24}{900} = \dfrac {121}{900}$ . Therefore, the solution is None of the others .