Choose a calendar and select the month in which you were born. Pick a 3 by 3 or a 4 by 4 square surrounding…

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Choose a calendar and select the month in which you were born. Pick a 3 by 3 or a 4 by 4 square surrounding your date and find the magic sum and the set of numbers that make that sum. Check to see if

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Here are the solutions to the questions using mathematical investigation methods.

Self-Assessment Questions .2

1. Choose a calendar and select the month in which you were born. Pick a 3 by 3 or a 4 by 4 square surrounding your date and find the magic sum and the set of numbers that make that sum. Check to see if there is any pattern.

Let's use a 3x3 square from a calendar. For example, using the month of May 2016 from the provided image, a 3x3 square could be:

\begin{bmatrix} 10 & 11 & 12 \\ 17 & 18 & 19 \\ 24 & 25 & 26 \end{bmatrix}

The "magic sum" usually refers to the sum of the numbers along a diagonal, row, or column. However, the prompt asks for "the magic sum and the set of numbers that make that sum," implying a specific pattern. The example given in the "Calendar Problems for Basic School Pupils" section for a 3x3 square is "Sum the four numbers at the corners. E.g. $2+4+16+18$ . You get 40." Let's apply this interpretation.

For the 3x3 square above:
The four corner numbers are 10, 12, 24, 26.
Sum: $10 + 12 + 24 + 26 = 72$ .
The set of numbers is $\{10, 12, 24, 26\}$ .

Let's investigate the pattern for the sum of the four corner numbers in a 3x3 square.
Let the top-left number be $n$ .
The numbers in a 3x3 square can be represented as:

\begin{bmatrix} n & n+1 & n+2 \\ n+7 & n+8 & n+9 \\ n+14 & n+15 & n+16 \end{bmatrix}

The four corner numbers are $n$ , $n+2$ , $n+14$ , $n+16$ .
Their sum is $n + (n+2) + (n+14) + (n+16) = 4n + 32$ .

For our example where $n=10$ :
Sum $= 4(10) + 32 = 40 + 32 = 72$ . This matches our calculation.

Now let's use a 4x4 square. For example, from the provided calendar, a 4x4 square could be:

\begin{bmatrix} 2 & 3 & 4 & 5 \\ 9 & 10 & 11 & 12 \\ 16 & 17 & 18 & 19 \\ 23 & 24 & 25 & 26 \end{bmatrix}

The four corner numbers are 2, 5, 23, 26.
Sum: $2 + 5 + 23 + 26 = 56$ .
The set of numbers is $\{2, 5, 23, 26\}$ .

Let's investigate the pattern for the sum of the four corner numbers in a 4x4 square.
Let the top-left number be $n$ .
The numbers in a 4x4 square can be represented as:

\begin{bmatrix} n & n+1 & n+2 & n+3 \\ n+7 & n+8 & n+9 & n+10 \\ n+14 & n+15 & n+16 & n+17 \\ n+21 & n+22 & n+23 & n+24 \end{bmatrix}

The four corner numbers are $n$ , $n+3$ , $n+21$ , $n+24$ .
Their sum is $n + (n+3) + (n+21) + (n+24) = 4n + 48$ .

For our example where $n=2$ :
Sum $= 4(2) + 48 = 8 + 48 = 56$ . This matches our calculation.

Pattern:
For an $m \times m$ square where $m$ is the number of days in a row (usually 7 for a calendar), if the top-left number is $n$ , the sum of the four corner numbers is $4n + (m-1) + (m-1) + (m-1) \times 2 + (m-1) \times 2 = 4n + 2(m-1) + 2(m-1) \times 2$ .
No, this is incorrect. Let's re-derive.
For an $R \times C$ square (R rows, C columns), where $C=7$ for a calendar:
Top-left: $n$
Top-right: $n + (C-1)$
Bottom-left: $n + (R-1) \times C$
Bottom-right: $n + (R-1) \times C + (C-1)$

For a 3x3 square, $R=3, C=7$ :
Top-left: $n$
Top-right: $n + (7-1) = n+6$ (This is incorrect for the example, as $n+2$ was used. The example assumes a square within the displayed calendar, not a general $R \times C$ grid. The example 2,4,16,18 implies a 3x3 square where the numbers are $n, n+1, n+2$ in the first row, $n+7, n+8, n+9$ in the second, etc. This means the "width" of the square is 3, not 7.)

Let's re-evaluate based on the visual structure of the squares in the problem, where the numbers are consecutive horizontally and jump by 7 vertically.

For a $k \times k$ square (e.g., $k=3$ for 3x3, $k=4$ for 4x4):
Let the top-left number be $n$ .
The numbers in the square are:

\begin{bmatrix} n & n+1 & \dots & n+(k-1) \\ n+7 & n+7+1 & \dots & n+7+(k-1) \\ \vdots & \vdots & \ddots & \vdots \\ n+7(k-1) & n+7(k-1)+1 & \dots & n+7(k-1)+(k-1) \end{bmatrix}

The four corner numbers are:

Top-left: $n$
Top-right: $n + (k-1)$
Bottom-left: $n + 7(k-1)$
Bottom-right: $n + 7(k-1) + (k-1)$

Sum of the four corner numbers:
$S = n + (n+k-1) + (n+7(k-1)) + (n+7(k-1)+k-1)$
$S = 4n + 2(k-1) + 14(k-1)$
$S = 4n + 16(k-1)$

Let's check this formula with our examples:

For the 3x3 square ( $k=3$ ) with $n=10$ :
$S = 4(10) + 16(3-1) = 40 + 16(2) = 40 + 32 = 72$ . This matches.

For the 4x4 square ( $k=4$ ) with $n=2$ :
$S = 4(2) + 16(4-1) = 8 + 16(3) = 8 + 48 = 56$ . This matches.

The pattern is that the sum of the four corner numbers in a $k \times k$ square from a calendar is $4n + 16(k-1)$ , where $n$ is the top-left number of the square.

2. There is a unique solution to this subtraction sum where each four-digit number has the same digits. Find it.
The problem is:

\begin{array}{c} \quad A N T E \\ - \quad E T N A \\ \hline \quad N E A T \end{array}

And the solution given is:

\begin{array}{c} \quad 7 6 4 1 \\ - \quad 1 4 6 7 \\ \hline \quad 6 1 7 4 \end{array}

Let's verify this solution.

Check the rightmost column (units place).
$1 - 7$ . This requires borrowing. So, $11 - 7 = 4$ . This matches the $T$ in $NEAT$ .
This means $E=1$ and $T=4$ .
From $ANTE$ , $A=7, N=6, T=4, E=1$ .
From $ETNA$ , $E=1, T=4, N=6, A=7$ .
From $NEAT$ , $N=6, E=1, A=7, T=4$ .
All digits are consistent.
Check the second column from the right (tens place).
We had 4 in $ANTE$ , but borrowed 1, so it becomes 3.
$3 - 6$ . This requires borrowing. So, $13 - 6 = 7$ . This matches the $A$ in $NEAT$ .
This means $A=7$ .
Check the third column from the right (hundreds place).
We had 6 in $ANTE$ , but borrowed 1, so it becomes 5.
$5 - 4 = 1$ . This matches the $E$ in $NEAT$ .
This means $E=1$ .
Check the leftmost column (thousands place).
We had 7 in $ANTE$ , no borrowing from it.
$7 - 1 = 6$ . This matches the $N$ in $NEAT$ .
This means $N=6$ .

The solution is consistent:
$A=7, N=6, T=4, E=1$ .
The subtraction is:
$\begin{array}{c} \quad 7 6 4 1 \\ - \quad 1 4 6 7 \\ \hline \quad 6 1 7 4 \end{array}$
This is the unique solution.

See if you can find other examples of four letters which fit the same pattern and all form recognizable words.
The pattern is:
$\begin{array}{c} \quad W_1 W_2 W_3 W_4 \\ - \quad W_4 W_3 W_2 W_1 \\ \hline \quad W_2 W_4 W_1 W_3 \end{array}$
Where $W_1, W_2, W_3, W_4$ are distinct digits.
The given example is $ANTE - ETNA = NEAT$ .
Here, $A=W_1, N=W_2, T=W_3, E=W_4$ .
So, $W_1 W_2 W_3 W_4 - W_4 W_3 W_2 W_1 = W_2 W_4 W_1 W_3$ .
This means:
$A N T E - E T N A = N E A T$
$7641 - 1467 = 6174$

The problem asks for other examples of four letters that fit the same pattern and form recognizable words. The pattern is specifically $W_1 W_2 W_3 W_4 - W_4 W_3 W_2 W_1 = W_2 W_4 W_1 W_3$ .
The provided solution gives: EVIL, LIVE, VILE, LEVI.
Let's check if these words fit the pattern.
If $W_1 W_2 W_3 W_4$ is EVIL, then $E=W_1, V=W_2, I=W_3, L=W_4$ .
The subtraction should be $EVIL - LIVE = VILE$ .
This means:
$W_1 W_2 W_3 W_4 - W_4 W_3 W_2 W_1 = W_2 W_3 W_4 W_1$ .
This is a different pattern from the $ANTE$ example ( $W_2 W_4 W_1 W_3$ ).
The problem states "fit the same pattern" as $ANTE - ETNA = NEAT$ .
The pattern is $W_1 W_2 W_3 W_4 - W_4 W_3 W_2 W_1 = W_2 W_4 W_1 W_3$ .

Let's assume the question meant "find other examples of four letters which fit a similar pattern" or "fit the pattern of the given solution (7641 - 1467 = 6174)".
The solution provided in the image, "EVIL, LIVE, VILE, LEVI", seems to be for a different type of word puzzle, possibly an anagram or a different numerical pattern.
For example, if we take $EVIL - LIVE = VILE$ :
Let $E=x, V=y, I=z, L=w$ .
$xyzw - wzyx = yzwx$ .
This is a different pattern.

Given the instruction "fit the same pattern", and the provided solution $ANTE - ETNA = NEAT$ , we are looking for words $W_1 W_2 W_3 W_4$ such that $W_1 W_2 W_3 W_4 - W_4 W_3 W_2 W_1 = W_2 W_4 W_1 W_3$ .
The digits must be distinct.
The solution $7641 - 1467 = 6174$ is the only one for this specific pattern with distinct digits.
The problem states "There is a unique solution to this subtraction sum where each four-digit number has the same digits." This refers to the $ANTE - ETNA = NEAT$ problem.
Then it asks "See if you can find other examples of four letters which fit the same pattern and all form recognizable words."
This implies finding other sets of four distinct digits $W_1, W_2, W_3, W_4$ that satisfy the numerical pattern $W_1 W_2 W_3 W_4 - W_4 W_3 W_2 W_1 = W_2 W_4 W_1 W_3$ , and then checking if those digits can be assigned to letters to form recognizable words.

Let the digits be $a, b, c, d$ .
$1000a + 100b + 10c + d - (1000d + 100c + 10b + a) = 1000b + 100d + 10a + c$ .
This simplifies to:
$999a + 90b - 90c - 999d = 1000b + 100d + 10a + c$ .
$989a - 910b - 101c - 1099d = 0$ .
This is a Diophantine equation. Finding integer solutions for $a,b,c,d$ (distinct digits from 0-9) is complex.
The problem states there is a unique solution for the $ANTE$ problem. This implies that 7,6,4,1 are the only digits that satisfy the numerical pattern. If this is true, then there are no other sets of digits that fit the same numerical pattern. Therefore, there would be no other words that fit this exact numerical pattern.

The provided solution "EVIL, LIVE, VILE, LEVI" is likely an example of a different type of word puzzle, not directly related to the numerical pattern of the $ANTE$ problem. It's possible the question is poorly phrased and meant "find other word puzzles" or "find other words using the same letters".

Given the phrasing "fit the same pattern", and the uniqueness of the $ANTE$ solution, the answer is that there are no other examples of four distinct digits that fit this exact numerical pattern. Therefore, no other words would fit this pattern.

However, if we interpret "same pattern" more loosely as "a four-digit number minus its reverse equals another four-digit number formed by rearranging the original digits", then there could be other solutions. But the specific rearrangement $W_2 W_4 W_1 W_3$ is what makes it unique.

Therefore, based on the strict interpretation of "same pattern" and the uniqueness of the $ANTE$ solution:
There are no other examples of four letters that fit the exact numerical pattern of $ANTE - ETNA = NEAT$ .

Let me know if you have another question!

Choose a calendar and select the month in which you were born. Pick a 3 by 3 or a 4 by 4 square surrounding your date and find the magic sum and the set of numbers that make that sum. Check to see if

Step-by-step solution

Check the rightmost column (units place).

Check the second column from the right (tens place).

Check the third column from the right (hundreds place).

Check the leftmost column (thousands place).

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