2015 Countdown Problem #15: Palindrome Integration

Algebra Level 5

Let x x be a real number and f ( x ) f(x) be a real-valued function on x x such that f ( x ) f(x) is:

  • the next palindrome larger than x x (if x x is not a palindrome); or

  • x x (if x x is a palindrome).

For example, f ( 1001 ) = 1001 f(1001)=1001 , f ( 1001.01 ) = f ( 1002 ) = 1111 f(1001.01)=f(1002)=1111 .

Let

A = 999 2015 f ( x ) d x A=\int _{ 999 }^{ 2015 }{ f(x) \mbox{ } dx }

B = f ( . . . ( f ( f ( 1603 ) 58 ) 58 ) . . . ) 58 B=f(...(f(f(1603)-58)-58)...)-58

Find the value of A B \frac{A}{B} .

This problem is part of the set 2015 Countdown Problems .


The answer is 990.

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1 solution

Omkar Kamat
Dec 24, 2014

f(1603)=1661. 1661- 58 =1603. So we get back to where we were. Iterating any number of times will get us back to 1603. So B=1603. Now for 999<x<=1001 we have f(x)= 1001. 1001<x<=1111 f(x)= 1111. Continuing in this fashion gives us f(x)= 2(1001) + 110 ( 1111+1221+ .... + 1991) + 11(2002) +13(2112) = 1586970 = A.

So A/B = 990

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