Box Progression Part 3

Logic Level 3

$\large \square + \square\square+\square\square\square+\square\square\square\square$

You are given that the numbers $0,1,2,\ldots,9$ are to be filled in the square boxes as shown above (without repetition) such that it represent a sum of a 1-digit, 2-digit, 3-digit, and 4-digit number.

Find total number of possible arrangements of these nine numbers such that the sum of these four numbers is minimized .

Details and Assumptions :

This is an arithmetic puzzle, where $1 \square$ would represent the 2-digit number 19 if $\square = 9$ . It does not represent the algebraic expression $1 \times \square$ .
The numbers are not allowed to have a leading 0 (even the 1-digit number).
See Part 1 , Part 2 and Part 4 .

The answer is 144.

1 solution

Satyen Nabar
Aug 17, 2015

The first two digits of $4$ - digit number are $10$ . The first digit of $3$ -digit number is $2$ . They are constant. The tens digit can be any of $3, 4, 5$ arranged in $3!= 6$ ways. The units digit can be any of $6, 7, 8, 9$ arranged in $4!=24$ ways. Total of $24\times 6=144$ numbers

I think this problem should be categorized in combinatorics and not in logic.

Patrick Engelmann - 5 years, 10 months ago

well.. maybe, but it's quite elementary combinatorics...

Nik Gibson - 2 years, 9 months ago

here in the question u have made a mistake... please specify that the number cannot start with a zero.... you have specified to only fill digits not that they have to be a real no. else the ans is 4! 3! 2! = 288

Anuj Modi - 5 years, 9 months ago

In the question, it's said that here is a 1- digit, a 2-digit, a 3-digit and a 4-digit number. So, 0 can't be at first.

Omar Sayeed Saimum - 4 years, 3 months ago

Box Progression Part 3

The answer is 144.

1 solution

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