(a) Choose a four-digit decimal number such that the outer digits sum to 10, the

Question

(a) Choose a four-digit decimal number such that the outer digits sum to 10, the inner digits sum to 8, and the second digit is one less than the first digit.

Now multiply this number by the last digit, and divide by the first digit. The result is the original number with the digits in reverse order. This is the digit-reversing property

(c) Use algebra to show why any four-digit number created according to the algorithm given in part (a) must have the digit-reversing property.

(d) Are there four-digit numbers in other bases with the digit-reversing property? Show that there are by developing an algorithm to generate them in base R, and use algebra to show that numbers generated by your algorithm will have the digit-reversing property. Then generate examples in binary, octal and hexadecimal and show that these numbers have the digit-reversing property.

Explanation / Answer

Interesting. with such a constraint on number choice, all other digits depend on the first one, if we called it x, the second digit is x-1, the third digit is 8-(x-1)=9-x, and the last digit is 10-x. Any of the 9 possible numbers made by this are therefore = 1000x+100*(x-1)+10*(9-x)+(10-x), which if you multiply it out is 1089x. If you multiply this by the last digit (10-x), and divide by the first digit (x) you get 1089*(10-x). The reverse of any of the original numbers is 1000*(10-x)+100*(9-x)+10*(x-1)+x which is 10890 -1089x = 1089*(10-x), so they are the same. For other bases, like octal it would just be the same idea, but with numbers like (8^3)*x+(8^2)*(x-1)+8*(7-x)+(8-x)=567x / x * (8-x) = 567*(8-x) Reversed origional: (8^3)*(x-8)+(8^2)*(7-x)+8*(x-1)+x= 567*(8-x) Ill leave generalizing the problem to base R up to you.

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