The all digit's sums

Probability Level 3

In how many different ways can the digits from 1 to 9 be placed into the squares (without repetition) so that the addition is valid?

If you need to use programming, it's OK. Please share your code.


The answer is 336.

Jose Fernandez Goycoolea by Jose Fernandez Goycoolea , 37, Chile

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2 solutions

Yuriy Kazakov
Aug 2, 2020 
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from itertools import permutations
s=0
for k in permutations([1,2,3,4,5,6,7,8,9],9):
  if (k[0]+k[3])*100+(k[1]+k[4])*10+k[2]+k[5] ==k[6]*100+k[7]*10+k[8]:
    s=s+1
print(s)


1
 
336

0 Helpful 0 Interesting 0 Brilliant 0 Confused

its 336 \boxed{336} , here is some code: 

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P = perms(1:9);
R =[];          
for k = 1:factorial(9)
    if 100*(P(k,1)+P(k,4)-P(k,7)) + 10*(P(k,2)+P(k,5)-P(k,8)) + (P(k,3)+P(k,6)-P(k,9)) == 0
        R = [R; P(k,:)] ;
    end
end
size(R,1)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Can we solve the problem without any computer program?Mr.(Goycoolea)!

Matin Naseri - 3 years, 2 months ago

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One could solve by inspection, finding essentially different solutions column by column and their permutations if possible, but it would be really tedious (you should take into account the possible carries).

Might be there is a cleaner way, but i haven't found any.

Jose Fernandez Goycoolea - 3 years, 2 months ago

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