98901 has an interesting property that if you reverse its digits (to form a different 5-digit number), then the ratio of the two numbers is an integer:
Likewise, there exists another positive 5-digit integer that has this property. What is it?
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The general formula for such number can be written as 9 9 ( 1 0 n − 1 ) − 1 1 ( 1 0 n − 1 ) = 8 8 ( 1 0 n − 1 ) = 8 ⋅ 1 1 ( 1 0 n − 1 ) , n ≥ 2
And hence the answer is 9 8 9 0 1 − 1 0 9 8 9 = 1 0 9 8 9 × 8 = 8 7 9 1 2 .
Generalization: Let us try to figure out the smallest possible number such × b 0 b n b 1 ⋯ b n − 1 b n − 1 ⋯ b 1 b n 9 b 0 For a single digit number 9 ⋅ b 0 = b 0 ⟹ b 0 = 0 . On the similar manner we can observe that for two and three digit number { 9 ( 1 0 b 0 + b 1 ) = 1 0 b 1 + b 0 ⇒ 8 9 b 0 = 9 0 b 1 9 ( 1 0 0 b 0 + 1 0 b 1 + b 2 ) = 1 0 0 b 2 + 1 0 b 1 + b 0 ⇒ 8 9 9 b 0 + 8 0 b 1 = 9 1 b 2 For the above two cases we cannot find such b 0 , b 1 , b 2 ∈ Z + < 1 0 . Trying for the 4 digit number we can observe that 8 9 9 9 b 0 + 8 9 0 b 1 − 1 0 b 2 − 9 9 1 b 3 = 0 Clearly b 0 = 1 implying b 3 = 9 from above cryptogram set and 8 9 0 b 1 − 1 0 b 2 = − 8 0 and hence we obtain b 1 = 0 , b 2 = 8 and 1 0 8 9 our smallest number.
Note that 1 1 ⋅ 9 9 = 1 1 ( 1 0 2 − 1 ) = 1 0 8 9 1 1 ⋅ 9 9 9 = 1 1 ( 1 0 3 − 1 ) = 1 0 9 8 9 1 1 ⋅ 9 9 9 9 = 1 1 ( 1 0 4 − 1 ) = 1 0 9 9 8 9 from the above pattern 1 1 ( 1 0 n − 1 ) , we can easily determine the smallest nth digit number for n ≥ 2 such property holds true. For 5 digit such number we set n = 3 giving us 1 0 9 8 9 . Reversing the digit we further observe that 9 8 0 1 9 8 9 0 1 9 8 9 9 0 1 = 9 9 ⋅ 9 9 = 9 9 ( 1 0 2 − 1 ) = 9 9 ⋅ 9 9 9 = 9 9 ( 1 0 3 − 1 ) = 9 9 ⋅ 9 9 9 9 = 9 9 ( 1 0 4 − 1 ) And we have then 9 9 ⋅ ( 1 0 n − 1 ) ÷ 1 1 ( 1 0 n − 1 ) = 9 . If we perform the subtraction we can observe that 9 9 ( 1 0 n − 1 ) − 1 1 ( 1 0 n − 1 ) = 8 8 ( 1 0 n − 1 ) = 8 ⋅ 1 1 ( 1 0 n − 1 ) set n = 2 , 3 , 4 , ⋯ then 8 ⋅ 1 1 ( 1 0 2 − 1 ) = 8 7 1 2 8 ⋅ 1 1 ( 1 0 3 − 1 ) = 8 7 9 1 2 8 ⋅ 1 1 ( 1 0 4 − 1 ) = 8 9 9 7 1 2 On reversing these numbers we can find the general formula 2 ⋅ 1 0 8 9 = 2 ⋅ 1 1 ( 1 0 n − 1 ) and hence 8 ⋅ 1 1 ( 1 0 n − 1 ) ÷ 2 ⋅ 1 1 ( 1 0 n − 1 ) = 4 .
I had set up a problem upon these amazing number few weeks ago here Reflection .