Box Progression Part 2

Logic Level 1

□ + □ □ + □ □ □ + □ □ □ □ \large \square + \square\square+\square\square\square+\square\square\square\square

Place each of the digits 0 through 9, without repetition, in the boxes above. Then what is the maximum possible sum?

Details and Assumptions :

  • This is an arithmetic puzzle, where 1 □ 1 \square would represent the 2-digit number 19 if □ = 9 \square = 9 . It does not represent the algebraic expression 1 × □ 1 \times \square .


The answer is 10656.

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Chung Kevin by Chung Kevin , 45, Australia

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15 solutions

Achille 'Gilles'
Sep 10, 2015

From the highest digit, start by filling the thousands, the hundreds, the tens and then, the units.

There is 288 (1! x 2! x 3! x 4!) possible arrangements as long as:

9 is at the thousands positions, [thousands x 9 = 9000]

8 and 7 are at the hundreds positions, [hundreds x (8 + 7) = 1500]

6, 5 and 4 are at the tens positions [tens x (6 + 5 + 4) = 150]

3, 2, 1 and 0 are at the units positions [units x (3 + 2 + 1 + 0) = 6]

So 9000 + 1500 + 150 + 6 = 10 656

47 Helpful 2 Interesting 2 Brilliant 2 Confused

hey its given not to repeat the digit

Manan Rastogi - 5 years, 6 months ago

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They aren't repeats. They are totals for each unit

Thomas Sergeant - 4 years, 11 months ago

This reminds me of chemistry - start filling the squares from the highest digit (first thousands then hundreds, the tens and last the units.)
So we get the following equation:
9852 + 743 + 61 + 0 = 10656

19 Helpful 1 Interesting 1 Brilliant 0 Confused

Thank you for your solution!

Chung Kevin - 5 years, 6 months ago
Oli Hohman
May 24, 2016

I will give my solution primarily in words. The way to look at this problem is to "put the most importance" on the highest digit (thousands digit). Therefore, the thousands digit will be 9.

Second in priority are the hundreds digits. There are two spots with hundreds digits, so just make sure you have 8 and 7 in either place.

The next thing to do is to analyze where you're going to want your lowest numbers. Since you want your highest numbers in the higher digits, you want your lower numbers in the lower digits. This means you should put 0,1, & 2 in the single digits in order to maximize the sum. From there you have:

9860+751+42+3 = 10656

The 2 and 3 can be interchanged without changing the sum.

12 Helpful 0 Interesting 0 Brilliant 0 Confused

You have repeated the 6 twice, the sum of your addition will be, 10676.

Satvik Tandon - 4 years, 11 months ago

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Edited. Thank you!

Oli Hohman - 4 years, 11 months ago

You are right in saying that the 2 and 3 can be interchanged, but you can extend that further to say that the numbers in the units value (i.e: 0, 1, 2, 3) can all be interchanged however one wishes as it will still give a unit sum of 6, so the sum remains unchanged.

Sylvester Mathiyalagan - 3 years, 3 months ago

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You're correct. Thanks.

Oli Hohman - 3 years, 2 months ago
Sadasiva Panicker
Sep 10, 2015

0+41+752+9863 =10656

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Way I did it but I did 40 + 1 only change

Luke Johnston - 5 years, 5 months ago
Ahsan Azhar
Sep 10, 2015

a) first fill the four digit highest place with higher number 9 ....b) then fill the 2nd highest digit of 4 digit number 2nd higher number 8 .... c) then fill 3 digit number hieghst digit place with 3rd higer number 7 ....d) then return to 4 digit number and fill the 3rd higehest digit place with 4rth higher digit 6.... e) then come to the 3 digit number and fill its middle digit with 5th higehr number 5 f) now come to 2 digits number and place 6th higher number 4 in its hieghts digit place ..... g) now come to the 4 digit number and place 7th higher number 3 to its last and minimum digit place ..... h) now come to 3 digit place and place 8th higher number 2 in its last and minimum digit place ..... i) now come to the 2 digit place and place 9th higher number 1 in its last and minimum place .... j) now we have only one number remaining 0 and only one remaining place is one digit number so place it there ..... now all arrangement is 0+41+752+9863 which is equal to 10656 which is maximum possible sum of this arrangement

3 Helpful 0 Interesting 0 Brilliant 0 Confused

how did you predict it? 0.o

silsila rahman - 5 years ago
Harsh Singh
Aug 15, 2015

You first have to maximize the thousands digit which is '9' at the 4digit number,then you maximize the hundredth digits which can be either at the 4 digit numbers or the 3 digit number by '8' or '7' then ,then you maximize the tenth digit by either '6' ,'5' or '4' and so on.

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Rishab Gupta
May 30, 2017

The pattern could be :- 9852 741 63 0 =10656

1 Helpful 0 Interesting 0 Brilliant 0 Confused
NotReal Mcgee
Apr 24, 2019

Biggest digits at the most significant positions

0 Helpful 0 Interesting 0 Brilliant 0 Confused 
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# https://brilliant.org/practice/arithmetic-puzzles-level-2-challenges/?p=4
from itertools import permutations
from math import isinf

perm = permutations([i for i in range(10)])

min_ = isinf
max_ = 0
max_permutation = ''
min_permutation = ''

for i in list(perm):
    a = int(i[0])
    b = int(str(i[1])+str(i[2]))
    c = int(str(i[3])+str(i[4])+str(i[5]))
    d = int(str(i[6])+str(i[7])+str(i[8])+str(i[9]))
    x = a + b + c + d
    if x < min_:
        min_ = x
        min_permutation = i
    if x > max_:
        max_ = x
        max_permutation = i


print 'Min: %d; %d + %d%d + %d%d%d + %d%d%d%d\n' %(min_, min_permutation[0], min_permutation[1], \
           min_permutation[2], min_permutation[3], min_permutation[4], min_permutation[5], \
           min_permutation[6], min_permutation[7], min_permutation[8], min_permutation[9])

print 'Max: %d; %d + %d%d + %d%d%d + %d%d%d%d\n' %(max_, max_permutation[0], max_permutation[1], \
           max_permutation[2], max_permutation[3], max_permutation[4], max_permutation[5], \
           max_permutation[6], max_permutation[7], max_permutation[8], max_permutation[9])

perm = permutations([i for i in range(10)])
min_count = 0
max_count = 0
for i in list(perm):
    a = int(i[0])
    b = int(str(i[1])+str(i[2]))
    c = int(str(i[3])+str(i[4])+str(i[5]))
    d = int(str(i[6])+str(i[7])+str(i[8])+str(i[9]))
    x = a + b + c + d
    if x == min_:
        min_count += 1
    if x == max_:
        max_count += 1

print 'There are %d number permutations possible to achieve total of %d\n' %(min_count, min_)
print 'There are %d number permutations possible to achieve total of %d' %(max_count, max_)


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Min: 450; 6 + 37 + 148 + 0259

Max: 10656; 0 + 41 + 752 + 9863

There are 288 number permutations possible to achieve total of 450

There are 288 number permutations possible to achieve total of 10656

0 Helpful 0 Interesting 0 Brilliant 0 Confused

The number of permutations is simply 1! * 2! * 3! * 4!

Francesco Chiariello - 2 years, 11 months ago

insert integers from given range based on order of length of squares, from left to right to return largest integer

xxxx 9760 xxx 851 xx 42 x 3 In [1]: 9760 + 851 + 42 + 3 Out[1]: 10656

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Easy one. If we look on the puzzle not horizontally buy vertically we start filling with the highest left digits the thousands hundreds etc. We get: 9863 752 41 0 = 10656

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Travell Criner
Feb 18, 2017

Let's label the digit in the ith box as a_i . We're constructing numbers in decimal in this problem. The sum we're constructing is:

a 1 + 10(a 2) + a 3 + 100(a 4) + 10(a 5) + a 6 + 1000(a 7) + 100(a 8) + 10(a 9) + a 10 = 1000(a 7) + 100(a 4 + a 8) + 10(a 2 + a 5 + a 9) + (a 1 + a 3 + a 6 + a 10) .

This is a polynomial in 10. Our sum is maximized when we first maximize the highest powers of 10 first. So then, we set our coefficient sums as follows:

a 7 = 9 a 4 + a 8 = 8 + 7 = 15 a 2 + a 5 + a 9 = 6 + 5 + 4 = 15 a 1 + a 3 + a 6 + a 10 = 3 + 2 + 1 + 0 = 6.

This has the effect of setting our polynomial as follows:

1000(9) + 100(15) + 10(15) + 6 = 10656.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Mc Riyuzaki
Jan 12, 2017

As simple as putting higher digits on higher places (thousands, hundreds, tens and ones) and avoid putting all higher numbers together in just one addends

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Adrian Adrian
Aug 25, 2015

Always place the highest numbers in the highest places which are the hundreds and the thousands and use the lower numbers at the lower points in the number, here is my solution.

9753 + 842 + 61 + 0

0 Helpful 0 Interesting 0 Brilliant 0 Confused

try to fill 9, 8, 7, ..., 1, 0 to the most maximum position of digits firstly

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you clarify the approach?

Chung Kevin - 5 years, 10 months ago

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9 8 6 3+7 5 2 + 4 1 + 0

9 will give thousand 8 , 7 will give 100 . . . . . . . .. .

maximum no.s will give maximum place value

Akash singh - 5 years, 10 months ago

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