9 Consecutive integers with answer 2 in a magic square
Answers
Solving an Odd-Numbered Magic Square
You can find this number by using a simple math formula, where n = the number of rows or columns in your magic square. So, for example, in a 3x3 magic square, n = 3. The magic constant = n[(n^2+1)/2]. So, in the example of the 3x3 square:
sum = 3 * [(9 + 1) / 2]
sum = 3 * (10 / 2)
sum = 3 * (5)
sum = 15
The magic constant for a 3x3 square is 15.
All rows, columns, and diagonals must add up to this number.
Place the number 1 in the center box on the top row. This is always where you begin when your magic square has odd-numbered sides, regardless of how large or small that number is. So, if you have a 3x3 square, place the number 1 in Box 2; in a 15x15 square, place the number 1 in Box 8.
Fill in the remaining numbers using an up-one, right-one pattern. You will always fill in the numbers sequentially (1, 2, 3, 4, etc.) by moving up one row, then one column to the right. You’ll notice immediately that in order to place the number 2, you’ll move above the top row, off the magic square. That’s okay — although you always work in this up-one, right-one manner, there are three exceptions that also have patterned, predictable rules:
If the movement takes you to a “box” above the magic square’s top row, remain in that box’s column, but place the number in the bottom row of that column.
If the movement takes you to a “box” to the right of the magic square’s right column, remain in that box’s row, but place the number in the furthest left column of that row.
If the movement takes you to a box that is already occupied, go back to the last box that has been filled in, and place the next number directly below it.
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A certain magic square contains 9 consecutive 2-digit numbers. The sum of numbers in any line is equal to one of the numbers in the square with digits reversed. This is still the case if 7 is added to each entry. What is the number in the center?
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Let ab be a 2 -digit number whose digits when reversed equals the sum of numbers in a line ( the magic constant ) .
If n is the middle term among nine consecutive terms, then the sum of all the nine terms is 9n , and the sum of each line is 3n .
Therefore, 10b+a=3n , where n is one of the nine consecutive integers one of which is ab . Hence, 3∣(10a+b) , and so 3∣(a+b) , and ∣∣10b+a3−(10a+b)∣∣≤4.…(1)
Inequality (1) simplifies to give ∣∣7b−29a∣∣≤12 . Thus, 29a≤7b+12≤(7⋅9)+12 , so that a∈{1,2} .
If a=1 , then 29–12≤7b≤29+12 , and so b=5 since 3∣(a+b) .
If a=2 , then 58–12≤7b≤58+12 , and so b=7 since 3∣(a+b) .
Thus, there are only two 2 -digit numbers that meet the given conditions: 15 and 27 . Note that 51=3⋅17 , 17∈{13,…,21} , and 72=3⋅27 , 27∈{20,…,28} . Incidentally, the second arithmetic progressions is obtained from the first by adding 7 to each term.
The number in the center is middle term of the first arithmetic progression, that is, 17 . ■