The 9 spots in a 3×3 grid are filled with distinct digits. Consider the 6 three-digit numbers obtained by reading left to right along rows, or down along columns. All 6 of these numbers are divisible by 6. How many are divisible by 5?
Second attempt at analytical solution.
Call the digits:
a b c
d e f
g h i
We know:
Each of {c,f,g,h,i} must each be even.
Each of {a,b,d,e} must each be odd, and are either {1,3,5,7} or {1,5,7,9}
Either way, modulo 3, {a,b,d,e} must be {0,1,1,2}
Each row sum and column sum must be 0 mod 3.
But the sum of a,b,d,e is 1 mod 3.
Considering the first two rows, c+f must be 2 mod 3 and therefore i must be 1 mod 3. The same reasoning applies to the first two columns.
Therefore i can never be 0 and there is always exactly 1 3-digit number divisible by 5. qed.
---- more verbose listing ----
Start with one possible arrangement (modulo 3) of {a,b,d,e}: {0,1,1,2}
This gives:
0 1 c
1 2 f
g h i
Then filling in the rest of the digits, mod 3:
0 1 2
1 2 0
2 0 1 <-- i=1
All 12 possible arrangements of a,b,d,e mod 3 all lead to i=1 mod 3.
The other 11 arrangements:
0 1 c
2 1 f --> i=1
g h i
0 2 c
1 1 f --> i=1
g h i
1 0 c
1 2 f --> i=1
g h i
1 0 c
2 1 f --> i=1
g h i
1 1 c
0 2 f --> i=1
g h i
1 1 c
2 0 f --> i=1
g h i
1 2 c
0 1 f --> i=1
g h i
1 2 c
1 0 f --> i=1
g h i
2 0 c
1 1 f --> i=1
g h i
2 1 c
0 1 f --> i=1
g h i
2 1 c
1 0 f --> i=1
g h i
Edited on January 19, 2025, 2:45 pm
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Posted by Larry
on 2025-01-19 14:17:00 |
Analytic solution
