Palindromes are forbidden

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:

98901 ÷ 10989 = 9. {98901}\div {10989} = 9 .

Likewise, there exists another positive 5-digit integer that has this property. What is it?


The answer is 87912.

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1 solution

Naren Bhandari
Nov 4, 2018

The general formula for such number can be written as 99 ( 1 0 n 1 ) 11 ( 1 0 n 1 ) = 88 ( 1 0 n 1 ) = 8 11 ( 1 0 n 1 ) , n 2 99(10^n-1) -11(10^n-1) = 88(10^n-1) = 8\cdot 11(10^n-1), \ \ n\geq 2

And hence the answer is 98901 10989 = 10989 × 8 = 87912 98901-10989 = 10989\times 8 =87912 .

Generalization: Let us try to figure out the smallest possible number such b 0 b 1 b n 1 b n × 9 b n b n 1 b 1 b 0 \begin{array} {ccccc} \large & & b_0 & b_1 \cdots b_{n-1} & b_n \\ \large \times & & & &9 \\ \hline \large & & b_n & b_{n-1} \cdots b_1 & b_0 &\end{array} For a single digit number 9 b 0 = b 0 b 0 = 0 9\cdot b_0 = b_0\implies b_0=0 . On the similar manner we can observe that for two and three digit number { 9 ( 10 b 0 + b 1 ) = 10 b 1 + b 0 89 b 0 = 90 b 1 9 ( 100 b 0 + 10 b 1 + b 2 ) = 100 b 2 + 10 b 1 + b 0 899 b 0 + 80 b 1 = 91 b 2 \begin{cases} 9(10b_0 + b_1) = 10b_1 +b_0 \Rightarrow 89b_0 = 90b_1\\ 9(100b_0+10b_1+b_2) = 100b_2+10b_1+b_0 \Rightarrow 899b_0 + 80b_1 =91b_2 \\ \end{cases} For the above two cases we cannot find such b 0 , b 1 , b 2 Z + < 10 b_0 ,b_1 ,b_2\in\mathbb Z^+ < 10 . Trying for the 4 digit number we can observe that 8999 b 0 + 890 b 1 10 b 2 991 b 3 = 0 8999b_0 + 890b_1 -10b_2 - 991b_3 =0 Clearly b 0 = 1 b_0=1 implying b 3 = 9 b_3=9 from above cryptogram set and 890 b 1 10 b 2 = 80 890b_1 -10b_2 =-80 and hence we obtain b 1 = 0 , b 2 = 8 b_1=0 ,b_2=8 and 1089 1089 our smallest number.

Note that 11 99 = 11 ( 1 0 2 1 ) = 1089 11 999 = 11 ( 1 0 3 1 ) = 10989 11 9999 = 11 ( 1 0 4 1 ) = 109989 \begin{aligned} 11\cdot 99= 11(10^2-1) =1089\\ 11\cdot 999=11(10^3-1) = 10989 \\11\cdot 9999 = 11(10^4-1)= 109989 \end{aligned} from the above pattern 11 ( 1 0 n 1 ) 11(10^n-1) , we can easily determine the smallest nth digit number for n 2 n\geq 2 such property holds true. For 5 digit such number we set n = 3 n=3 giving us 10989 10989 . Reversing the digit we further observe that 9801 = 99 99 = 99 ( 1 0 2 1 ) 98901 = 99 999 = 99 ( 1 0 3 1 ) 989901 = 99 9999 = 99 ( 1 0 4 1 ) \begin{aligned} 9801 & =99\cdot 99 = 99(10^2-1) \\ 98901 & = 99\cdot 999= 99(10^3-1) \\ 989901 & =99\cdot 9999 = 99(10^4-1) \end{aligned} And we have then 99 ( 1 0 n 1 ) ÷ 11 ( 1 0 n 1 ) = 9. 99\cdot(10^n-1) ÷ 11(10^n-1) = 9. If we perform the subtraction we can observe that 99 ( 1 0 n 1 ) 11 ( 1 0 n 1 ) = 88 ( 1 0 n 1 ) = 8 11 ( 1 0 n 1 ) 99(10^n-1) -11(10^n-1) = 88(10^n-1) = 8\cdot 11(10^n-1) set n = 2 , 3 , 4 , n= 2,3,4,\cdots then 8 11 ( 1 0 2 1 ) = 8712 8 11 ( 1 0 3 1 ) = 87912 8 11 ( 1 0 4 1 ) = 899712 \begin{aligned} 8\cdot 11(10^2-1) = 8712 \\ 8\cdot 11(10^3-1) = \boxed{87912}\\ 8\cdot 11(10^4-1) = 899712 \end{aligned} On reversing these numbers we can find the general formula 2 1089 = 2 11 ( 1 0 n 1 ) 2\cdot 1089 = 2\cdot 11 (10^n-1) and hence 8 11 ( 1 0 n 1 ) ÷ 2 11 ( 1 0 n 1 ) = 4. 8\cdot 11(10^n-1) ÷ 2\cdot 11(10^n-1)=4.

I had set up a problem upon these amazing number few weeks ago here Reflection .

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