Not as hard as it seems

Find a 10 10 digit number made from all 10 10 digits 0 9 0-9 such that the number made from the first 2 2 digits is divisible by 2 2 , from the first 3 3 is divisible by 3 3 and so on.


The answer is 3816547290.

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1 solution

James Moors
Apr 23, 2015

Firstly, I eliminated possibilities for each digit of a b c d e f g h i j \overline{abcdefghij} :

j = 0 e = 5 j=0 \Rightarrow e=5

( b , d , f , h ) ( 2 , 4 , 6 , 8 ) ( a , c , g , i ) ( 1 , 3 , 7 , 9 ) (b,d,f,h) \in (2,4,6,8) \Rightarrow (a,c,g,i) \in (1,3,7,9)

4 a b c d 4 c d 4|\overline{abcd} \Rightarrow 4|\overline{cd} , so c ( 1 , 3 , 7 , 9 ) d ( 2 , 6 ) c \in (1,3,7,9) \Rightarrow d \in (2,6)

8 a b c d e f g h 8 f g h 8|\overline{abcdefgh} \Rightarrow 8|\overline{fgh} , so f ( 2 , 4 , 6 , 8 ) , g ( 1 , 3 , 7 , 9 ) h ( 2 , 6 ) f \in (2,4,6,8), g \in (1,3,7,9) \Rightarrow h \in (2,6)

Finally, ( d , h ) ( 2 , 6 ) ( b , f ) ( 4 , 8 ) (d,h) \in (2,6) \Rightarrow (b,f) \in (4,8) .

Then, I made lists of each n n -digit number which was valid:

a b ( 14 , 18 , 34 , 38 , 74 , 78 , 94 , 98 ) \overline{ab} \in (14, 18, 34, 38, 74, 78, 94, 98)

a b c ( 147 , 183 , 189 , 381 , 387 , 741 , 783 , 789 , 981 , 987 \overline{abc} \in (147, 183, 189, 381, 387, 741, 783, 789, 981, 987

I'll skip a few lines, because d d just tacks ( 2 , 6 ) (2,6) onto each and e e sticks on a 5 5 , so each set has 20 20 elements apiece.

a b c d e f ( 147258 , 183654 , 189654 , 381654 , 387654 , 741258 , 783654 , 789654 , 981654 , 987654 ) \overline{abcdef} \in (147258, 183654, 189654, 381654, 387654, 741258, 783654, \\789654, 981654, 987654)

a b c d e f g ( 1472583 , 3816547 , 7836549 ) \overline{abcdefg} \in (1472583, 3816547, 7836549)

a b c d e f g h = 38165472 a b c d e f g h i j = 3816547290 \overline{abcdefgh} = 38165472 \Rightarrow \overline{abcdefghij} = 3816547290

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