Grids & Bubbles

Logic Level 2

The above is a 3 × 3 3\times 3 table, where each cell consists of a distinct digit from 1 to 9 with 7 in the center.

The circular bubbles on the borderlines indicate the absolute differences between the pair of numbers in the neighboring cells, 8 of which are known and colored yellow, and 4 of which--A, B, C, D--are unknown and colored red.

What is the 4-digit integer A B C D ? \overline{ABCD}?


The answer is 2161.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Calvin Lin Staff
Dec 28, 2016

It seems like there isn't enough information to get started, since we're only given the absolute differences. There is likely some case checking involved, but we want to minimize the number of cases. Certainly, looking at 8 ! 8! cases isn't ideal.

Which part of the puzzle contains the most information? If the absolute difference of 2 numbers is high, then this means that the numbers are very restrictive (must be very large and very small). Hence, let's focus our attention on the bottom middle (BM) and bottom left (BL) squares, which have a difference of 6. The possible pairs are: 1 7 , 2 8 , 3 9 1-7, 2-8, 3-9 .

Let's label the squares by their row (Top / Center / Bottom) and their column (Left / Middle / Right).

Case 1: B M = 1 BM = 1 .
Then B L = 7 BL = 7 which contradicits the distinct digits condition.

Case 2: B M = 2 BM = 2 .
Then B R = 7 BR = 7 , which contradicts the distinct digits condition.

Case 3: B M = 3 BM = 3 .
Then B L = 9 , B R = 8 BL = 9, BR = 8 .
Then C L = 5 , C R = 6 CL = 5, CR = 6 .
But then TR would be 3 or 9, which contradicts the distinct digits condition.

Case 4: B M = 7 BM = 7 .
This contradicts the distinct digits condition.

Case 5: B M = 8 BM = 8 .
Then B L = 2 , B R = 3 BL = 2, BR = 3 .
Then C L = 6 CL = 6 .
Then T L = 5 TL = 5 (since it is not 7).
The digits we have left are 1, 4, 9. Since T R T M = 5 , T R C R = 3 |TR - TM| = 5, |TR - CR| = 3 , hence we must have T R = 4 , T M = 9 , C R = 3 TR = 4, TM = 9, CR = 3 .

Case 6: B M = 9 BM = 9
Then B L = 3 , B R = 4 BL = 3 , BR = 4 .
Then C L = 7 CL = 7 , which contradicts the distinct digits condition.

In conclusion, only case 5 is possible, and we get A B C D = 2161 \overline{ABCD} = {2161} .

Precisely, there are only at most 4 cases to try given that 7 is out already. :)

Worranat Pakornrat - 4 years, 5 months ago

Log in to reply

Agreed. We just differ at the stage of ruling out the 1 7 1-7 case. I left it in to show that at the start, I was only using information from the C M C L = 6 |CM - CL| = 6 observation and nothing else.

Calvin Lin Staff - 4 years, 5 months ago
Daanish Bansal
Jan 6, 2017

just check 8 possible cases for placing 9 and now as 9 is largest no. we know the two numbers in its neigbourhood and thus every case proceeds smoothly

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...