Playing With Numbers

Step 1: Look at the unit's column: $A + 5 = 2$ (or ends in $2$, meaning $12$).

$$A + 5 = 12 \implies A = 12 - 5 = \mathbf{7}$$

Since the sum is $12$, write $2$ and carry over $1$ to the ten's column.
Step 2: Look at the ten's column with the carryover of $1$:

$$1 + 3 + 2 = B \implies B = \mathbf{6}$$

• Final Answer:

$$\mathbf{A = 7, \quad B = 6}$$

Check: $37 + 25 = 62$. Correct!

• Write the numbers in generalised place value form:

$$abc = 100a + 10b + c$$

$$cba = 100c + 10b + a$$

• Subtract $cba$ from $abc$:

$$abc - cba = (100a + 10b + c) - (100c + 10b + a)$$

$$= 100a - a + 10b - 10b + c - 100c$$

$$= 99a - 99c$$

$$= \mathbf{99(a - c)}$$

• Since $99(a - c)$ has $99$ as an explicit factor, the difference is always divisible by $99$ (and consequently also by $9$ and $11$).

• By the Divisibility Rule for $9$, a number is divisible by $9$ if and only if the sum of its digits is a multiple of $9$.
• Calculate the sum of the digits:

$$\text{Sum} = 2 + 4 + x + 5 = 11 + x$$

• Since $x$ is a single digit ($0 \le x \le 9$), the possible multiples of $9$ greater than or equal to $11$ are $18$ and $27$.
• Setting $11 + x = 18$:

$$x = 18 - 11 = \mathbf{7}$$

(Setting $11 + x = 27$ gives $x = 16$, which is not a single digit).
• Final Answer:

$$\mathbf{x = 7}$$

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• A number is divisible by $3$ if the sum of its digits is divisible by $3$.

$$\text{Sum} = 3 + 1 + y + 5 = 9 + y$$

• For $(9 + y)$ to be a multiple of $3$, $y$ can be any single digit such that $y$ is a multiple of $3$ ($0 \le y \le 9$).
• Possible values of $y$:

$$\mathbf{y = 0, \quad y = 3, \quad y = 6, \quad y = 9}$$

• There are $4$ possible values for $y$.

• Label digit positions from right to left (units is position 1):
Number: $1 \; 0 \; 8 \; 2 \; 4$
• Digits at odd places (1st, 3rd, 5th): $4, 8, 1$

$$\text{Sum of odd place digits } (S_{\text{odd}}) = 4 + 8 + 1 = 13$$

• Digits at even places (2nd, 4th): $2, 0$

$$\text{Sum of even place digits } (S_{\text{even}}) = 2 + 0 = 2$$

• Calculate the difference:

$$\text{Difference} = S_{\text{odd}} - S_{\text{even}} = 13 - 2 = \mathbf{11}$$

• Since the difference ($11$) is a multiple of $11$, the number $10824$ is completely divisible by $11$.

Step 1: In the unit's place, $B \times 3$ ends in $B$.
Testing digits for $B$:
• $0 \times 3 = 0$ (ends in $0 \implies B = 0$)
• $5 \times 3 = 15$ (ends in $5 \implies B = 5$)
Step 2: If $B = 0$: $A0 \times 3 = CA0$. Divide by 10: $A \times 3 = CA$. No non-zero single digit satisfies this.
Step 3: If $B = 5$:

$$A5 \times 3 = CA5$$

$5 \times 3 = 15$ (write $5$, carry $1$).
In ten's column: $(A \times 3) + 1$ must end in $A$.
Test digits for $A$:
• If $A = 2$: $(2 \times 3) + 1 = 7$ (does not end in $2$).
• If $A = 5$: $A$ cannot equal $B$.
• If $A = 7$: $(7 \times 3) + 1 = 22$ (does not end in $7$).
• If $A = 0$: Leading digit cannot be 0.
Wait, test $50 \times 3 = 150$: Here $A = 5, B = 0, C = 1$! Check: $50 \times 3 = 150$ ($AB \times 3 = CAB$).
• Final Answer:

$$\mathbf{A = 5, \quad B = 0, \quad C = 1}$$

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• Let the repeated digit be $a$ ($1 \le a \le 9$).
The number is $aaa = 100a + 10a + a = 111a$.
• Factorise the number $111$:

$$111 = 3 \times 37$$

$$aaa = 3 \times 37 \times a$$

• Since $3$ and $37$ are explicit prime factors in $111a$, any such number is always divisible by both $3$ and $37$ (e.g., $777 \div 37 = 21, 777 \div 3 = 259$).

Apply the Magic Constant formula: $M = \frac{n(n^2 + 1)}{2}$.
Here $n = 4$:

$$M = \frac{4(4^2 + 1)}{2} = \frac{4(16 + 1)}{2} = \frac{4(17)}{2} = 2 \times 17 = \mathbf{34}$$

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In a $4 \times 4$ magic square, every row, column, and diagonal sums to $34$.