## Problem Of The Week 25

**Let X and Y be distinct 3 digit palindromes such that X>Y. How many pairs (X,Y) exist such that X-Y is also a 3 digit palindrome ?**

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**Let X and Y be distinct 3 digit palindromes such that X>Y. How many pairs (X,Y) exist such that X-Y is also a 3 digit palindrome ?**

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Written by Implex

September 24, 2008 at 1:53 am

Posted in Combinatorics, Number Thoery, Problem of the week

Tagged with Combinatorics, Number Theory, Problem of the week

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Have you given the solution anywhere?

RiddhiSeptember 24, 2008 at 12:13 pm

I give solutions after people write their attempts !

Post your attempt.

outtimedSeptember 24, 2008 at 12:37 pm

My answer is 1620.

Suppose x=100*x1+10*x2+x1 =101*x1+10*x2

y=100*y1+10*y2+y1 =101*y1+10*y2

Hence x-y= 101*(x1-y1)+10*(x2-y2)

10*(x2-y2) will end in zero.

For any value of x1-y1, resultant x-y will be a palindrome if and only if x2>y2.

For example if (x1-y1)=2, 101*(x1-y1)=202. Suppose x2-y2= 6, x-y=262, another palindrome. On the other hand, if x-y=-6, x-y=202-60=142, which is not a palindrome.

For x1-y1=1, there are 8 pairs of x1,y1 (2,1),(3,2),(4,3),(5,4),(6,5),(7,6),(8,7) and (9,8). Corresponding to each value of x1-y1, there will be 45 pairs of (x2,y2)such that x2>y2.

Similarly for x1-y1=2, there are 7 pairs of (x1,y1) with corresponding 45 pairs of (x2,y2).

So on and so forth, for x1-y1=8, there is only one pair of (x1,y1) with corresponding 45 pairs of (x2,y2).

Hence total number of ordered pairs of x and y is

45*(8+7+6+5+4+3+2+1) = 45*36 = 1620

Girish

GirishSeptember 24, 2008 at 4:26 pm

you have done a minor error

try again girish

answer is 1980

outtimedSeptember 24, 2008 at 6:05 pm

hey Girish great solutions but the errors i think is that its not 36*45 ie but 36*55 because the middle number of both x and y can be the same ,hence 45 + 10 ( because of 10 digits ).

aceSeptember 24, 2008 at 7:48 pm

Let the two palindromes be

aba=X

cdc=Y

Their difference is ede

Now, for ede to be 3 digit, a>1 and c can take values from 1 upto a-1

Now, we see that the difference is a palindrome only if the middle digit of papa(aba)is >=middle digit of munna(cdc)

so, for a=2, c=1

and b can take values from 0 to 9.

For, b=0, d=0(1 value)

b=1, d=0/1(2 values)

so on.. we get the possible no. values =1+2+3+4+5+6+7+8+9+10=55

Now, for a=3, c=1/2

similarly possible values are

a=4, c=1/2/3

a=5, c=1/2/3/4

a=6, c=1/2/3/4/5

a=7, c=1/2/3/4/5/6

a=8, c=1/2/3/4/5/6/7

a=9, c=1/2/3/4/5/6/7/8

so, in all a can take 1+2+3+4+5+6+7+8=36

So, the number of possible numbers is 36*55

milindSeptember 25, 2008 at 5:16 pm

nice work milind, but there is not need to list down all

just follow the pattern and

you will get it.

good job

outtimedSeptember 25, 2008 at 10:16 pm