Compute 2018

Computer Science Level 2

Find the highest 9 9 -digit number a b c d e f g h i ‾ \overline{abcdefghi} with distinct non-zero digits a , b , c , d , e , f , g , h , a, b, c, d, e, f, g, h, and i , i, such that a ( b + c ( d + e ( f + g ( h + i ) ) ) ) = 2018. a(b+c(d+e(f+g(h+i)))) = 2018. and that a , c , e , a, c, e, and g g are not equal to 1 1 .


The answer is 298546371.

Rajen Kapur by Rajen Kapur , 72, India

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2 solutions

Giorgos K.
May 15, 2018

using M a t h e m a t i c a Mathematica

FromDigits/@Select[Permutations@Range@9,#[[1]](#[[2]]+#[[3]](#[[4]]+#[[5]](#[[6]]+#[[7]](#[[8]]+#[[9]]))))==2018&]

returns ALL possible numbers

{128754369,128754396,128934756,128934765,135798624,135798642,164782539,164782593,183492657,183492675,213847569,213847596,214936578,214936587,245968317,245968371,287536419,287536491,295864713,295864731,298546317,298546371}

so we start searching for the highest that meets the criteria
starting from the last one and... it is indeed the right one!

298546371 \boxed{298546371}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Zeeshan Ali
May 15, 2018 
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import itertools # https://docs.python.org/2/library/itertools.html#itertools.permutations
import numpy as np # http://www.numpy.org/

# set of all the possible 9 non-zero-digit numbers in decending order
permutations = list(itertools.permutations([i for i in range(9, 0, -1)], 9))

# for every p = (a, b, c, d, e, f, g, h, i) in above set
for p in permutations:
    # if all elements in p are unique
    if len(np.unique(p)) == 9:
        # if a*(b+c*(d+e*(f+g*(h+i)))) equals 2018
        if p[0]*(p[1]+p[2]*(p[3]+p[4]*(p[5]+p[6]*(p[7]+p[8])))) == 2018:
            # print the elements or digits of that number 'p'
            print (p)
            # since the p's in permutations are in decending order therefore the first one satisfying
            # the condition is the larget such number, hence break (i.e. stop further search)
            break

(2, 9, 8, 5, 4, 6, 3, 7, 1)

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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