Almost Primes Square

Algebra Level 5

You're given a 4 × 4 4\times 4 magic square, where the sum of every row, column, or diagonal equals to 180, and all the letters A A to O O represent distinct 2-digit prime numbers apart from 49 as shown.

The unit numbers never repeat in any sum, and the corner numbers are arranged such that the product A O L C AOLC is maximum.

If A A > O O > L L > C C , what is the value of A A + B B - C C + D D - E E + F F - G G + H H - I I + J J - K K + L L - M M + N N - O O ?


The answer is 45.

Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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2 solutions

If we add the sums of the two diagonals and four edges together, we will get:

6 × 180 6\times 180 = 3( A A + C C + L L + O O ) + ( B B + D D + E E + F F + G G + H H + I I + J J + K K + M M + N N )

Subtracting the sum of four rows (A to O) from the sums above, we will get:

( 6 − 4 ) × 180 (6-4)\times 180 = 360 = 2( A A + C C + L L + O O )

Thus, 180 = A A + C C + L L + O O .

That means the sum of the corner numbers also equals to 180, and from comparing the sum of corner numbers and the diagonals, we will get:

A A + C C + L L + O O = A A + E E + J J + O O = C C + F F + I I + L L

C C + L L = E E + J J and A A + O O = F F + I I

Therefore, A A + C C + L L + O O = E E + F F + I I + J J = 180.

That means, not only the corner numbers add up to 180, but also the central ones!

Also, C C - E E = J J - L L and A A - F F = I I - O O , restricting the gaps along the knight-move orders.

Since all the alphabetical variables are all primes, to get equal amount in each row, column, or diagonal, the ending digits 1, 3, 7, 9 should be spread equally among the squares. Here is one example:

...1 ...7 49 ...3 ...9 ...3 ...1 ...7 ...3 ...9 ...7 ...1 ...7 ...1 ...3 ...9

With one known number 49, all we need to do is to plug in the digits that fit the restrictions above, and we know that the sum of the 4 digits will equal to 16 because 180 - (1+3+7+9) = 160.

With trial & error, we will reach a conclusion that the gaps between the corner and central numbers will equal to 2, and the solution magic square is shown below:

As a result, A A + B B - C C + D D - E E + F F - G G + H H - I I + J J - K K + L L - M M + N N - O O = 61 +47 - 23 + 29 - 43 + 41 - 67 + 53 - 79 + 17 - 31 + 37 - 11 + 73 - 59 = 45.

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Very sorry my answer does not have distinct prime as required. \color{#D61F06}{\text{Very sorry my answer does not have distinct prime as required.}}

The missing total in the 1st column is 180 − 49 = 131. 131 3 = 43 2 3 . / t e x t P r i m e n e a r 43 a r e 41 , 43 , 47 , a n d 41 + 43 + 47 = 131. So using these numbers and 49 so that no number appears twice in any row , column or diagonal, we get the above square. 180 - 49 = 131.~~ \dfrac {131} 3 =43\frac 2 3.\\ /text{Prime near 43 are 41 , 43 , 47 , and 41 + 43 + 47 =131.} \\ \text {So using these numbers and 49 so that no number appears twice in any}\\ \text{row , column or diagonal, we get the above square.}\\~~\\

1st column is plus. Last minus. They add up to zero.
Second column 180 minus 47 =133 is minus.
That 47 plus third column less 49=47+180-49=180-2=178 is plus.
So required sum is 178 - 133= 45 \Huge ~~~~~~~~~\color{#D61F06}{45}


0 Helpful 0 Interesting 0 Brilliant 0 Confused

Actually there's way you can use distinct primes for this magic square (no repeated numbers) though you've got the right answer. ;)

Worranat Pakornrat - 5 years, 5 months ago

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Sorry, I again missed the point. You are right. The way I got the total may be easier.

Niranjan Khanderia - 5 years, 5 months ago

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