Question

Let $A = \{n \in N: n \text{ is 3-digit number}\}$, $B = \{9k + 2: k \in N\}$ and $C = \{9k + l: k \in N\}$ for some $l(0 < l < 9)$. If the sum of all the elements of the set $A \cap (B \cup C)$ is $274 \times 400$, then $l$ is equal to

• A. 5
• B. 7
• C. 9
• D. 20

Answer 1

Answer: (A)

5

Answer 2

Let $A = \{n \in N: n \text{ is 3-digit number}\}$. So, $A = \{100, 101, ..., 999\}$. $A$ is an arithmetic progression with first term $a_1 = 100$, last term $a_m = 999$, and common difference $d=1$. The number of elements in $A$ is $999 - 100 + 1 = 900$.

Let $B = \{9k + 2: k \in N\}$. Since $k \in \{1, 2, 3, ...\}$, the elements of $B$ are $\{11, 20, 29, ...\}$. These are natural numbers that leave a remainder of 2 when divided by 9.

Let $C = \{9k + l: k \in N\}$ for some $l(0 < l < 9)$. Since $k \in \{1, 2, 3, ...\}$ and $l \in \{1, 2, ..., 8\}$, the elements of $C$ are $\{9+l, 18+l, 27+l, ...\}$. These are natural numbers that leave a remainder of $l$ when divided by 9.

We are interested in the set $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$. The sum of elements in this set is given as $274 \times 400 = 109600$.

First, let's find the elements in $A \cap B$. These are 3-digit numbers of the form $9k+2$. The smallest 3-digit number is 100. We need $9k+2 \ge 100 \implies 9k \ge 98 \implies k \ge 98/9 \approx 10.89$. Since $k \in N$, the smallest integer $k$ is 11. The smallest term is $9(11)+2 = 101$. The largest 3-digit number is 999. We need $9k+2 \le 999 \implies 9k \le 997 \implies k \le 997/9 \approx 110.78$. Since $k \in N$, the largest integer $k$ is 110. The largest term is $9(110)+2 = 992$. The set $A \cap B = \{101, 110, 119, ..., 992\}$. This is an arithmetic progression with first term $a_1 = 101$, common difference $d=9$. To find the number of terms, let $992 = 101 + (n_B - 1)9$. $891 = (n_B - 1)9 \implies n_B - 1 = 99 \implies n_B = 100$. The sum of elements in $A \cap B$ is $S_{A \cap B} = \frac{n_B}{2}(a_1 + a_{n_B}) = \frac{100}{2}(101 + 992) = 50(1093) = 54650$.

Next, let's find the elements in $A \cap C$. These are 3-digit numbers of the form $9k+l$, where $l \in \{1, 2, ..., 8\}$. The smallest 3-digit number is 100. We need $9k+l \ge 100 \implies 9k \ge 100-l \implies k \ge (100-l)/9$. Since $l \in \{1, ..., 8\}$, $100-l$ is between 92 and 99. $(100-l)/9$ is between $92/9 \approx 10.22$ and $99/9 = 11$. For any $l \in \{1, ..., 8\}$, the smallest integer $k \ge (100-l)/9$ is 11. The smallest term is $9(11)+l = 99+l$. The largest 3-digit number is 999. We need $9k+l \le 999 \implies 9k \le 999-l \implies k \le (999-l)/9$. Since $l \in \{1, ..., 8\}$, $999-l$ is between 991 and 998. $(999-l)/9$ is between $991/9 \approx 110.11$ and $998/9 \approx 110.89$. For any $l \in \{1, ..., 8\}$, the largest integer $k \le (999-l)/9$ is 110. The largest term is $9(110)+l = 990+l$. The set $A \cap C = \{99+l, 108+l, ..., 990+l\}$. This is an arithmetic progression with first term $a'_1 = 99+l$, common difference $d=9$. The values of $k$ range from 11 to 110. The number of terms is $n_C = 110 - 11 + 1 = 100$. The sum of elements in $A \cap C$ is $S_{A \cap C} = \frac{n_C}{2}(a'_1 + a'_{n_C}) = \frac{100}{2}((99+l) + (990+l)) = 50(1089 + 2l)$.

Now, consider the intersection $(A \cap B) \cap (A \cap C)$. This set contains 3-digit numbers that are in both $B$ and $C$. A number is in $B$ if it leaves a remainder of 2 when divided by 9. A number is in $C$ if it leaves a remainder of $l$ when divided by 9. A number is in both sets if it leaves a remainder of 2 and $l$ when divided by 9. This is possible only if $l=2$.

Case 1: $l \ne 2$. If $l \ne 2$, then a number cannot leave remainders 2 and $l$ simultaneously when divided by 9. So, $B \cap C = \emptyset$. Thus, $(A \cap B) \cap (A \cap C) = \emptyset$. The sum of elements in $A \cap (B \cup C) = S_{(A \cap B) \cup (A \cap C)} = S_{A \cap B} + S_{A \cap C} - S_{(A \cap B) \cap (A \cap C)} = S_{A \cap B} + S_{A \cap C} - 0$. The sum is $54650 + 50(1089 + 2l) = 54650 + 54450 + 100l = 109100 + 100l$. We are given that the sum is $109600$. $109100 + 100l = 109600$ $100l = 109600 - 109100 = 500$ $l = 5$. This value $l=5$ is in the range $0 < l < 9$ and $l \ne 2$. So, $l=5$ is a possible solution.

Case 2: $l = 2$. If $l=2$, then $C = \{9k+2: k \in N\} = B$. In this case, $A \cap (B \cup C) = A \cap (B \cup B) = A \cap B$. The sum of elements is $S_{A \cap B} = 54650$. The given sum is $109600$. Since $54650 \ne 109600$, $l$ cannot be 2.

The only possible value for $l$ is 5.

The final answer is $\boxed{5}$.

Explanation of the solution:

1. Identify the elements of sets A, B, and C based on their definitions.
2. Determine the elements of $A \cap B$, which are 3-digit numbers of the form $9k+2$. Find the range of k for these numbers and calculate the sum of the arithmetic progression.
3. Determine the elements of $A \cap C$, which are 3-digit numbers of the form $9k+l$. Find the range of k for these numbers and calculate the sum of the arithmetic progression in terms of $l$.
4. Determine the elements of $(A \cap B) \cap (A \cap C)$. These are 3-digit numbers that leave a remainder of 2 and $l$ when divided by 9. This is only possible if $l=2$.
5. Consider two cases: $l \ne 2$ and $l = 2$.
6. If $l \ne 2$, the intersection $(A \cap B) \cap (A \cap C)$ is empty. The sum of elements in $A \cap (B \cup C)$ is the sum of elements in $A \cap B$ plus the sum of elements in $A \cap C$. Equate this sum to the given value and solve for $l$.
7. If $l = 2$, the set $C$ is the same as set $B$. The set $A \cap (B \cup C)$ is $A \cap B$. Calculate the sum and check if it matches the given value.
8. The case $l=2$ does not match the given sum. The case $l \ne 2$ yields $l=5$, which is a valid value for $l$ in the specified range.