Question

If number of seven digit numbers in which first four digits are in ascending order, last four digits are in descending order and middle digit is a perfect square of natural number, is $\lambda$, then digit at the tens place of $\lambda$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Answer 2

Let the seven-digit number be $d_1 d_2 d_3 d_4 d_5 d_6 d_7$.

The digits $d_1, d_2, \dots, d_7$ are from the set $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. Since it is a seven-digit number, the first digit $d_1$ cannot be 0. So $d_1 \in \{1, 2, \dots, 9\}$.

The first four digits are in ascending order: $d_1 < d_2 < d_3 < d_4$. Since $d_1 \ge 1$, all the first four digits $d_1, d_2, d_3, d_4$ must be non-zero. Thus, $1 \le d_1 < d_2 < d_3 < d_4$. This implies that $d_1, d_2, d_3$ must be distinct digits greater than or equal to 1 and less than $d_4$. The smallest possible values for $d_1, d_2, d_3$ are 1, 2, 3. Thus, $d_4$ must be at least 4.

The last four digits are in descending order: $d_4 > d_5 > d_6 > d_7$. This implies that $d_4, d_5, d_6, d_7$ must be distinct. The digits $d_5, d_6, d_7$ must be distinct digits less than $d_4$. The digits can include 0. The smallest possible distinct values for $d_5, d_6, d_7$ are 0, 1, 2. Thus, $d_4$ must be at least 3.

The middle digit is $d_4$. It is given that $d_4$ is a perfect square of a natural number. Natural numbers are $1, 2, 3, \dots$. Their squares are $1^2=1, 2^2=4, 3^2=9, 4^2=16, \dots$. Since $d_4$ is a single digit, $d_4$ must be from $\{0, 1, \dots, 9\}$. So, $d_4$ must be 1, 4, or 9.

Combining the conditions on $d_4$:

1. $d_4 \ge 4$ (from $d_1 < d_2 < d_3 < d_4$ and $d_1 \ge 1$).
2. $d_4 \ge 3$ (from $d_4 > d_5 > d_6 > d_7$).
3. $d_4 \in \{1, 4, 9\}$.

From these conditions, the possible values for $d_4$ are 4 and 9.

Case 1: $d_4 = 4$. The first four digits: $d_1 < d_2 < d_3 < 4$. Since $d_1 \ge 1$, the digits $d_1, d_2, d_3$ must be chosen from $\{1, 2, 3\}$. There is only $\binom{3}{3}=1$ way to choose these three distinct digits, which are $\{1, 2, 3\}$. Since they must be in ascending order, $(d_1, d_2, d_3) = (1, 2, 3)$. So the first four digits are (1, 2, 3, 4). The last four digits: $4 > d_5 > d_6 > d_7$. The digits $d_5, d_6, d_7$ must be chosen from $\{0, 1, 2, 3\}$. There are $\binom{4}{3}=4$ ways to choose three distinct digits from this set. Once chosen, there is only one way to arrange them in descending order. So, for $d_4=4$, the number of possible seven-digit numbers is $1 \times \binom{4}{3} = 1 \times 4 = 4$.

Case 2: $d_4 = 9$. The first four digits: $d_1 < d_2 < d_3 < 9$. Since $d_1 \ge 1$, the digits $d_1, d_2, d_3$ must be chosen from $\{1, 2, \dots, 8\}$. There are $\binom{8}{3}$ ways to choose three distinct digits from this set. Once chosen, there is only one way to arrange them in ascending order. $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$. The last four digits: $9 > d_5 > d_6 > d_7$. The digits $d_5, d_6, d_7$ must be chosen from $\{0, 1, \dots, 8\}$. There are $\binom{9}{3}$ ways to choose three distinct digits from this set. Once chosen, there is only one way to arrange them in descending order. $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$. So, for $d_4=9$, the number of possible seven-digit numbers is $\binom{8}{3} \times \binom{9}{3} = 56 \times 84$. $56 \times 84 = 56 \times (80 + 4) = 56 \times 80 + 56 \times 4 = 4480 + 224 = 4704$.

The total number of such seven-digit numbers, $\lambda$, is the sum of the numbers from Case 1 and Case 2. $\lambda = 4 + 4704 = 4708$.

We are asked to find the digit at the tens place of $\lambda$. $\lambda = 4708$. The digits of $\lambda$ are: 4 (thousands place), 7 (hundreds place), 0 (tens place), 8 (units place). The digit at the tens place of $\lambda$ is 0.