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).
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 ).
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
Have you given the solution anywhere?
ReplyDeleteI give solutions after people write their attempts !
ReplyDeletePost your attempt.
My answer is 1620.
ReplyDeleteSuppose 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
you have done a minor error
ReplyDeletetry again girish
answer is 1980
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 ).
ReplyDeleteLet the two palindromes be
ReplyDeleteaba=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
nice work milind, but there is not need to list down all
ReplyDeletejust follow the pattern and
you will get it.
good job