Question

Five different digits from the set of numbers {1, 2, 3, 4, 5, 6, 7} are written in random order. Then, the number of numbers that can be formed using 5 different digits from this set if the number is divisible by 9, is

Answer 1

240

Answer 2

A number is divisible by 9 if the sum of its digits is divisible by 9.

The given set of digits is $S = \{1, 2, 3, 4, 5, 6, 7\}$. The sum of all digits in the set $S$ is $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$.

We need to form 5-digit numbers using 5 different digits from this set such that the number is divisible by 9. First, we need to select a subset of 5 distinct digits from the set $S$ whose sum is divisible by 9.

Let the two digits from $S$ that are not chosen be $x$ and $y$, where $x, y \in S$ and $x \neq y$. The sum of the 5 chosen digits is $28 - (x + y)$.

For this sum to be divisible by 9, $28 - (x + y)$ must be a multiple of 9. Since $28 \equiv 1 \pmod{9}$, we have $1 - (x + y) \equiv 0 \pmod{9}$. This implies $x + y \equiv 1 \pmod{9}$.

We need to find pairs of distinct digits $\{x, y\}$ from the set $\{1, 2, 3, 4, 5, 6, 7\}$ such that their sum $x + y$ has a remainder of 1 when divided by 9. The minimum possible sum of two distinct digits from $S$ is $1 + 2 = 3$. The maximum possible sum of two distinct digits from $S$ is $6 + 7 = 13$. We are looking for $x + y$ in the range $[3, 13]$ such that $x + y \equiv 1 \pmod{9}$. The possible values for $x + y$ are $1, 10, 19, \dots$. Within the range $[3, 13]$, the only possibility is $x + y = 10$.

We need to find distinct pairs $\{x, y\}$ from $\{1, 2, 3, 4, 5, 6, 7\}$ such that $x + y = 10$. The possible pairs are:

• $3 + 7 = 10$. The pair is $\{3, 7\}$. Both 3 and 7 are in $S$.

• $4 + 6 = 10$. The pair is $\{4, 6\}$. Both 4 and 6 are in $S$.

So, the possible pairs of distinct digits that are left out are $\{3, 7\}$ and $\{4, 6\}$. This means the set of 5 digits used to form the number must be $S \setminus \{3, 7\}$ or $S \setminus \{4, 6\}$.

Case 1: The set of 5 digits is $S \setminus \{3, 7\} = \{1, 2, 4, 5, 6\}$. The sum of these digits is $1 + 2 + 4 + 5 + 6 = 18$, which is divisible by 9. Any 5-digit number formed using these 5 distinct digits will have a sum of digits equal to 18, and hence will be divisible by 9. The number of distinct 5-digit numbers that can be formed using these 5 digits is the number of permutations of 5 distinct objects, which is $5!$. $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

Case 2: The set of 5 digits is $S \setminus \{4, 6\} = \{1, 2, 3, 5, 7\}$. The sum of these digits is $1 + 2 + 3 + 5 + 7 = 18$, which is divisible by 9. Any 5-digit number formed using these 5 distinct digits will have a sum of digits equal to 18, and hence will be divisible by 9. The number of distinct 5-digit numbers that can be formed using these 5 digits is the number of permutations of 5 distinct objects, which is $5!$. $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

The total number of 5-digit numbers that can be formed using 5 different digits from the set $\{1, 2, 3, 4, 5, 6, 7\}$ such that the number is divisible by 9 is the sum of the numbers from Case 1 and Case 2. Total number = $120 + 120 = 240$.