Addition

Logic Level 1

$\large \begin{array} { rrrrr } &\square& \square & \square & \square & \square \\ + &\square& \square & \square & \square & \square \\ \hline \end{array}$

The above represents addition of two 5-digit integers. Each box represents a distinct, single-digit, non-negative integer. What is value of the final sum?

66666 77777 88888 99999

2 solutions

Marta Reece
May 30, 2017

Sum of all the digits from $0$ to $9$ is $45$ .

$45$ distributed evenly into $5$ places is $X=\dfrac{45}{5}=\boxed{9}$

One such solution is:

$43210+56789=99999$

And there is no solution with a different value of $X$ .

There could conceivably be one if we could carry $1$ , once or multiple times.

Carrying $1$ will drop the overall sum by $9$ , as we have $10$ less in one column and a $1$ added to the column in front of it.

Subracting $9$ , up to four times, from $45$ will give us sums of $36, 27, 18$ , or $9$

None of these numbers is divisible by $5$ , so there is no possibility of a second solution.

But what if we have carry over?

Calvin Lin Staff - 4 years ago

Viki Zeta
May 30, 2017

The main clue in the question is,

Each digit is distinct, containing numbers from 0 to 9 (inclusive)

Therefore, if any one box contains 0 , the resultant X vertically below it must contain the number in the very vertical box w.r.t 0 .

All numbers from 0 to 9, when added in a pair, must give the same number. This is possible only with the highest number in the group 9 .

ie,

But what if we have carry over? I agree that 9 is a possible answer. How do we know that there are no other solutions?

E.g. we could have $0 + 5 = 5$ and $9 + 6 = 15$ . (Of course, as it turns out, we can't fill out the grid.

Calvin Lin Staff - 4 years ago

Addition

2 solutions

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