## A.9 Digit Problems

1. Use $$3,4,5,6,7,8$$ to make digits between 5,000 & 6,000.
2. 2, 3, 4, 5, 6, 7: 6-digit numbers not divisible by 5
3. 5, 6, 7, 8, 0: Five digit numbers divisible by 4.
4. Make 7-digit numbers using 3, 4, 5, 5, 3, 4, 5, 6, keeping odd digits at odd positions, without using digits more than its frequency.
5. Use 1, 2, 3, 4, 5, 6, 7, 8, 9 to make numbers with even digits at beginning and end, using each digit only one.
6. Form numbers with 0, 3, 5, 6, 8 greater than 4000, without repeating any digit.
• 4-digits and starts with 5 $$\rightarrow 1 \times \space ^5P_3$$
• □ □ □ □ □ □ $$\rightarrow 5! \times \space ^5P_1$$ or 6!-5!
• Last two: 56,68, 76 and 60, 08, 80 $$\rightarrow 3! \times ^3P_1 + ^2P_1 \times 2! \times ^3P_1=18+12=30$$
• Odd positions = 4, even = 3;
there are repetitions. $$\rightarrow \frac {4!}{2!2!} \times \frac{3!}{2!}=18$$
• $$^4P_1 \times ^3P_1 \times 7! = 60480$$ or $$^4P_2 \times 7!$$
• □ □ □ □ + □ □ □ □ □ $$\rightarrow ^3P_1 \times ^4P_3 + ^4P_1 \times 4! = 168$$

1. Make meaningful odd numbers using the digits 6, 5, 2, 3, 0 once in each number.
2. Make meaningful even numbers using 5, 3, 2, 6, 0.
3. Use 1, 2, 3, 4 to make 3 or less-digit numbers by using digits more than once/any no. of times
• $$^2P_1 \times ^3P_1 \times 3!=36$$
• 2 at end, 6 at end, 0 at end $$\rightarrow 2 (1 \times 4!-3!)+4!$$
• 1-digit + 2-digit+3-digit (□ □ □) $$\rightarrow ^4P_1+4 \times 4 + 4^3$$