Christmas Streak 22/88: Four Square Theorem

Logic Level 4

French mathematician Joseph-Louis Lagrange proved that an arbitrary natural number can be expressed as the sum of four squares of non-negative integers, in 1772, which is known as the famous " Four Square Theorem ."

Let's prove it ourselves, then!

Solve the cryptogram below, and submit your answer as the 6-digit $\overline{\color{#E81990}N\color{#CEBB00}U\color{#624F41}M\color{#BBBBBB}B\color{#69047E}E\color{#3D99F6}R}.$

$\begin{array}{cccccccc} &&\color{#D61F06}S&\color{#EC7300}Q&\color{#CEBB00}U&\color{#20A900}A&\color{#3D99F6}R&\color{#69047E}E \\ &&\color{#D61F06}S&\color{#EC7300}Q&\color{#CEBB00}U&\color{#20A900}A&\color{#3D99F6}R&\color{#69047E}E \\ &&\color{#D61F06}S&\color{#EC7300}Q&\color{#CEBB00}U&\color{#20A900}A&\color{#3D99F6}R&\color{#69047E}E \\ +&&\color{#D61F06}S&\color{#EC7300}Q&\color{#CEBB00}U&\color{#20A900}A&\color{#3D99F6}R&\color{#69047E}E \\ \hline &&\color{#E81990}N&\color{#CEBB00}U&\color{#624F41}M&\color{#BBBBBB}B&\color{#69047E}E&\color{#3D99F6}R \end{array}$

This problem is a part of <Christmas Streak 2017> series .

The answer is 596328.

1 solution

Edwin Gray
Sep 26, 2018

Since the sum of the 4 S's does not result in 2 digits, S must equal to 1 or 2. Experimenting with RE, by trying values for E, it soon becomes apparent that the only combination that fits is R= 8, E = 2. Then we know that S = 1.It would seem that the presence of 0 might prove fruitful. Indeed, by trial and error, we find that A = 0, will result in B = 3. Trying values for U,we find that 9 works with Q = 4. Ed Gray

Christmas Streak 22/88: Four Square Theorem

The answer is 596328.

1 solution

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