Box Progression Part 4

Logic Level 2

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

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 ten numbers such that the sum of these four numbers is maximized .

Details and Assumptions :

For the purposes of this question, 0 is considered a 1-digit number.
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 answer is 288.

1 solution

Satyen Nabar
Aug 17, 2015

The first digit of $4$ - digit number is $9$ . That is constant. The hundreds digit can be any of $7, 8$ arranged in $2!= 2$ ways. . The tens digit can be any of $4, 5, 6$ arranged in $3!= 6$ ways. The units digit can be any of $0, 1, 2, 3$ arranged in $4!=24$ ways. Total of $2\times24\times 6=288$ numbers

Again, I think this is rather combinatorics than logic.

Patrick Engelmann - 5 years, 10 months ago

Can 0 be considered a single digit number?

Manoj Gunzalo Raj V - 5 years, 8 months ago

Units digit can not be 0. Problem says that "..possible arrangements of these nine numbers such that the sum of these four numbers is maximized.". Best case scenario, using 0 instead of 1 will result in a max sum of 10655 and the max possible sum is 10656. So, 0 can not be cosidered. There are 2!x3!x3! = 2x6x6 = 72 numbers that maximize the sum.

Enzo Amario - 4 years, 10 months ago

Every number must be used once. 9860 + 751 + 42 + 3 = 10655. therefore the solution is correct there are 2! 3! 4! ways to make maximize the sum

Andrew Good - 3 years, 2 months ago

I think the solution 2!3!4! is wrong. You can think of 4! combinations or 2ways times 3ways times 4ways. Both give 24 combinations. But I don't think there is 2!ways times 3! ways times 4! ways.

harry Juan - 2 years, 12 months ago

I think the solution 2!3!4! is wrong. You can think of 4! combinations or 2ways times 3ways times 4ways. Both give 24 combinations. But I don't think there is 2!ways times 3! ways times 4! ways.

harry Juan - 2 years, 12 months ago

Second digit has 2! ways,third digit has 3! ways,fourth digit has 4! ways.2!3!4! is correct.

X X - 2 years, 11 months ago

First places are occupied by 0,1,2,3 they are arranged in 4! Ways and second places by 6,5,4 they are arranged in 3! Ways and third place by 8,7 they are arranged in 2! ways and finaly at 4th place by 9 and arranged in 1! Ways. Finally total arrangements are 4! 3! 2!*1!=288

android 17 - 2 years, 2 months ago

Box Progression Part 4

The answer is 288.

1 solution

0 pending reports