Question

Let X be a two digit number such that both X and X2 end with the same digit and none of the digits in X equals zero. When the digits of X are written in the reverse order, the square of the new number so obtained has last digit as 6 and is less than 3000. Then the number of distinct possibilities for X is given by

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

Let $X$ be a two-digit number represented as $10a + b$, where $a$ is the tens digit and $b$ is the units digit. The problem states that $a, b \in \{1, 2, \dots, 9\}$.

Condition 1: $X$ and $X^2$ end with the same digit. The last digit of $X$ is $b$. The last digit of $X^2$ is the last digit of $b^2$. This implies $b \equiv b^2 \pmod{10}$. Checking values for $b \in \{1, 2, \dots, 9\}$:

• $1^2 = 1 \implies 1 \equiv 1 \pmod{10}$ (Valid)
• $5^2 = 25 \implies 5 \equiv 5 \pmod{10}$ (Valid)
• $6^2 = 36 \implies 6 \equiv 6 \pmod{10}$ (Valid) Thus, possible values for $b$ are $\{1, 5, 6\}$.

Condition 2: Let $Y$ be the number obtained by reversing the digits of $X$, so $Y = 10b + a$. $Y^2$ ends with 6 and $Y^2 < 3000$. The last digit of $Y$ is $a$. The last digit of $Y^2$ is the last digit of $a^2$. This implies $a^2 \equiv 6 \pmod{10}$. Checking values for $a \in \{1, 2, \dots, 9\}$:

• $4^2 = 16 \implies 4 \equiv 6 \pmod{10}$ (Valid)
• $6^2 = 36 \implies 6 \equiv 6 \pmod{10}$ (Valid) Thus, possible values for $a$ are $\{4, 6\}$.

Now we combine the possible values for $a$ and $b$ and check the condition $Y^2 < 3000$.

Case 1: $a=4$

• If $b=1$: $X = 41$. $Y = 14$. $Y^2 = 14^2 = 196$. Since $196 < 3000$, $X=41$ is a valid possibility.
• If $b=5$: $X = 45$. $Y = 54$. $Y^2 = 54^2 = 2916$. Since $2916 < 3000$, $X=45$ is a valid possibility.
• If $b=6$: $X = 46$. $Y = 64$. $Y^2 = 64^2 = 4096$. Since $4096 \not< 3000$, $X=46$ is not valid.

Case 2: $a=6$

• If $b=1$: $X = 61$. $Y = 16$. $Y^2 = 16^2 = 256$. Since $256 < 3000$, $X=61$ is a valid possibility.
• If $b=5$: $X = 65$. $Y = 56$. $Y^2 = 56^2 = 3136$. Since $3136 \not< 3000$, $X=65$ is not valid.
• If $b=6$: $X = 66$. $Y = 66$. $Y^2 = 66^2 = 4356$. Since $4356 \not< 3000$, $X=66$ is not valid.

The distinct valid possibilities for $X$ are 41, 45, and 61. Therefore, the number of distinct possibilities for $X$ is 3.