**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?

CommentsThus 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