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# 2018 NSW: Maths Puzzle 3

3rd puzzle by Dr Harry Wiggins

A palindromic number is a positive integer (consisting of one or more digits) that remains the same when the digits are reversed. For example: 386683,2442 and 111 (Nelson number, see https://en.wikipedia.org/wiki/111_(number)) are palindromic numbers, but 3447 and 200 are not. There is exactly one palindromic number with the following property: if you add 2018 to it, the result is again a palindromic number. What is this number?

7/31/2018 8:36 PM
First this palindrome, p, must be at least 111, as the minimum difference between two 4 digit or more palindromes, is 110, and the lowest palindrome greater, than 110 is 111. The palindrome must also be smaller than 1001, as if it is a 4-digit palindrome, it must be of the form 1001a+110b, for digits a and b. Thus 1001a+110b+2018 must equal another palindrome, of the form 1001c+110d, for some digits c and d.
Thus 1001a +2002+16+110b=1001c+110d
therefore 1001(a+2)+110b+16=1001c+110d
therefore 1001(a+2-c)+16=110(d-b) , but since d-b >=9, this cannot be true.
Therefore p is a 3 digit palindrome.
Thus p is of the form 101a+10b, for digits a and b.
Thus 101a+10b+2018=1001c+110d
Thus 101a +10b +2002+16=1001c+101d+10d-d
Thus 101(a-d)+10(b-d)=1001(c-2)-d-16
if a-d and b-d are both positive, then the LHS is a 3 digit palindrome, but that is not possible as 1001(c-2)-d-16 is either a 4 digit number, or a number from 975 to 984.
Thus it must be equal to 979, but that means d must equal both zero (a-d=9) and 5 (1001-16-5 = 979)
(c-2) cannot be less than -1, as The RHS would become too small. Thus c-2=0, therefore c=2. (a-d) cannot be less than -1, as the LHS would become too small. Thus a-d =0 implying a=d.
We now have
10(b-d)=-d-16
thus 10b-10d=-d-16
thus 10b-9d=-16
thus 9d-10b=16
thus 9d-10(b+1)=6,
thus 9d ens in a 6,
thus d=4,
thus b=2
thus  a=4
thus the palindrome is 4(101)+2(10)=424
Sachin Reddy Northwood School
8/1/2018 3:04 PM
Well done Sachin!!!
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