Palin - Palin

Let $A=\overline{a_1a_2a_2a_1},B=\overline{b_1b_2b_2b_1},C=\overline{c_1c_2c_1}$ ,while $A-B<1000$ ,it immediately imply that $a_1-b_1=1$ or $0$ . We cannot have $a_1-b_1=0$ ,because this imply $C=\overline{0c_20}$ which is clearly not a palindrome as the conditions given.Then $a_1-b_1=1$ ,and $C=\overline{1c_21}$ .Now, $11|A,B$ gives $11|A-B=C$ . Only $c_2=2$ satisfy $11|\overline{1c_21}$ (as $1-c_2+1=11k$ for some nonnegative integer $k,c_2<10$ ,then $c_2=2$ ), which gives $C=121$ . The required answer is $121$ .

Since C is the difference between A and B and C is a 3 digit number, the difference between A and B is less than 1000.

We consider 2 cases:

Case 1: The Thousands digits of A and B are the same.

We let $A=\overline{abba}$ and $B=\overline{acca}$ . Then $A-B=\overline{bb0}-\overline{cc0}=110(b-c)=C$ . Also, $1 \le b-c \le 9$ . Thus C is a 3 digit multiple of 110. All of such numbers are in the form of $\overline{aa0}$ with $1 \le a \le 9$ . There are clearly no palindromes that take that form.

Case 2: The Thousands digit of A and B are difference.

It is clear that the thousands digit of A must be one more than the thousands digit of B.

We write $A=\overline{(a+1) bb (a+1)}$ and $B=\overline{acca}$ . Then $A-B=1001+\overline{bb0}-\overline{cc0}=1001+110(b-c)=C$ . Since $C<1000$ , $b-c < 0$ . However, since C is at least 100, $c-b$ is at most 8.

Note that $C=1001-110(c-b)=1001-110c+110b=1+10(100+11b-11c)=1+10(100-11(c-b))$ .

Hence C could be 1+10(100-11k) where k ranges from 1 to 8. Thus C could be one of 891, 781, 671, 561, 451, 341, 231, and 121.

However, we are given that C is a palindrome, so we know that $C=\boxed{121}$ .

let A='abba', B='cddc'. Since C>0 and does not have 0 in the one's place we know a>c. Also since C is only three digits, d>b.

Then we set C='xyx'

x=a-c

y=10+b-d

and the palindrom condition:

x=10a+(b-1)-10c-d

we eliminate a,b,c,d from the last equation by substitution and get

y=11-9x

the only solution with x,y, integers less than 9 is x=1, y=2

so C=121

Let a,b,c,and d represent digits where A = acca and B = bddb.

Since A-B produces a three digit number, leaving zero in the 1000's place, we suspect that a = b, yielding: acca - bddb = 0??0

But that leaves zero in the one's place, which couldn't result in a three digit palindrome. The only other way to get a zero in the 1000s place is if a = b + 1 and d > c leaving C = 0??1. Or rather, because C is a palindrome, we get C = 1e1 for some digit 'e'.

In the process of subtracting the digits in the 10's and 100's place we note that the result in the 100's place must be one less than in the 10's place because we did not borrowed from the 10's place but we did borrow from the 100's place. That leaves us with C = 121.

Let $A = abba$ , a 4 digit number, and $B = cddc$ .

Consider A - B, Case 1 : $a = c$ This leads to a solution of $(0)(b-d)(b-d)(0)$ going according to the thousands,hundreds,tens and ones place. As one can see, a 3-digit number with the first 2 digit equal, and the ones place having 0 is never a palindrome number.

Case 2: $a = c + 1, b > d$ The subtraction would yield $(1)(b-d)(b-d)(1)$ , in which is a 4-digit number

Case 3(The only case left!) $a = c + 1, b < d$ Subtraction would yield $(0)(b+9-d)(b+10-d)(1)$ . In this case, there is a possibility of a palindrome number present. For the number to be a palindrome number, the ones place = hundreds place. Hence, $b + 9 - d = 1$ => $d - b = 8$ . Hence, the only possible integer value of $d$ and $b$ is $9$ and $1$ respectively. This hence implies that the palindrome number is $(1+9-9)(1+10-1)(1)$ => $121$

As A and B are 4-digit palindromes, there exist integers a, b, c, and d such that \begin{align} 1 &\le a,c \le 9 \tag{1} \\ 0 &\le b,d \le 9 \tag{2} \end{align} and \begin{align} A &= 1001a + 110b \\ B &= 1001c + 110d. \end{align} Then, $A - B = 1001(a-c) + 110(b-d).$ Let x = a - c and y = b - d . Then $C = A - B = 1001x + 110y.$ As A - B = C and C is positive, it must be the case that x is non-negative. From (1) and (2) we conclude that \begin{align} 0 \le x \le 8 \\ -9 \le y \le 9. \end{align} Recall that every positive integer can be written uniquely in the form $a_0 10^0 + a_1 10^1 + a_2 10^2 + ... + a_n 10^n$ where $0 \le a_0, a_1, ..., a_n \le 9.$

Suppose y is non-negative. Then, C can be written uniquely as $C = 1000x + 100y + 10y + x.$ As C is a 3-digit palindrome, it must be the case that x=0 and x=y . But then C=0 , which is a contradiction. We conclude that y is negative.

Now, C can be written uniquely in the form $C = 1000(x-1) + 100(9 + y) + 10(10 + y) + x. \tag{3}$ As C is a 3-digit palindrome, it must be the case that x-1=0 , and thus x=1 . It also must be the case that x = 9 + y , and thus y = x - 9 = -8 . Substituting these values in (3) gives $C = 121.$

We are given that: $C = (a_1-b_1)\times1000 + (a_2-b_2)\times100 + (a_2-b_2)\times10 + (a_1-b_1)$ , where $C$ is a three digit palindromic number.

We can immediately rule out the possibility of $a_1 = b_1$ , since this would imply $C = (a_2-b_2)\times100 + (a_2-b_2)\times10 = (a_2-b_2)\times110$ , which cannot be palindromic. Therefore, in order for $C$ to be a three digit number, it must be the case that $a_1 - b_1 = 1$ . This gives us the following information:

$C = 1001 + (a_2-b_2)\times110$

The values of $a_2-b_2$ that maintain $C$ as a three digit number are $-1, -2, ... , -9$ . The only value for which $C$ is a palindromic number is $-8$ , in which case $C = 1001 + (-8)\times110 = 121$ .

[Note: I am not very content with that last step, where I search over all possible values of $a_2 - b_2$ . I hope someone has a more intelligent solution for this final step.]

So we are looking for: $\overline{abba}-\overline{cddc}=\overline{efe}$ for which $a,c,e$ are not 0.

$\overline{abba}-\overline{cddc}=\overline{efe}$ $1000a+100b+10b+a-(1000c+100d+10d+c)=100e+10f+e$ $1001(a-c)+110(b-d)=101e+10f$ $11(91(a-c)+10(b-d))=101e+10f$

Now this means $101e+10f$ is a multiple of 11. Additionally from: $1001(a-c)+110(b-d)=101e+10f$ We see $a-c=1$ , this is so as;

If $a-c\geq 2 \Rightarrow 1001(a-c)+110(b-d)\geq 2002-990=1012$ , which is not possible as the result will be 4 digit.

If $a-c=0$ then $110(b-d)=101e+10f$ , which is not possible as $110(b-d)$ cannot be a 3 digit palindrome number for $(b-d)=0,1,2,3,4,5,6,7,8,9$

If $a-c\leq -1 \Rightarrow 1001(a-c)+110(b-d) \leq -1001+990=-11$ which is not possible as it has to be positive.

So as $(a-c)=1$ this means $e=1$ . So the only number that is palindrome with its first and unit digit 1, and is a multiple of 11, is 121.

4-digit palindrome numbers are necessarily multiples of 11. More specifically, they are of the form $11 * aba$ , where $aba$ is a 3-digit palindrome (note that not all 3-digit palindromes work). So $A - B$ can be written as $11(100(a - c) + 10(b - d) + (a - c)) = 11(101(a - c) + 10(b - d))$

The only 3-digit palindrome numbers that are also multiples of $11$ are those which are a multiple of $11^2$ . Thus $101(a -c) + 10(b - d)$ is a multiple of 11 less than 100. This is only possible when it equals 11. Thus, $C = 11 x 11 = 121$ .

If we take 2 palindromes such that the first and last digits of them have a difference of 2, then their difference will be more than 999. Lets say that A = abba, and B = cddc (the letters represent the digits).

We know that a - c = 1 (that's what I conclude on the first paragraph). Then the first and third digits of the 3 digit palindrome is 1. Note that a and c also needs to appear at the end of the number (left extreme of both numbers), so we need to substract in such a way that when we get to the a, we had already regrouped.

If b > d, the regrouping is impossible. Same happens if b = d. So b < d, because it's the only way of regrouping. When we substract b - d, we need to take from the other b, so that we get b + 10 - d = x (we dont know x yet). Then on the other b, we will substract b - 1 - d, because we took from that b. But b still being minor than d, so when we regroup we will have b + 9 - d, and we know b + 9 - d = 1. So we get b + 10 - d = x and b + 9 - d = 1, what clearly means that x = 2, so C = 121.

Note: Sorry that my english is not so fluent. :D

Let the subtraction statement be:

abba

cddc

= efe

the letters represent digits. from this, we can already determine two statements.

d > b because if we were to single out bb - dd = ef, then there needs to be carrying a one at some point to produce ef. if b was greater than d, then if you subtracted it, it should equal repeating digits instead of ef.

a - 1 = c because in the thousands place, it is 0 because we are given that a four dig number minus a four dig number equals a three dig. in this case, we know that the hundreds place will take away 1, so that a = c. therefore, a minus 1 equals to c.

now that we know that, we know that e is automatically equal to 1 because if a minus c, that will equal to 1. for this entire question, it doesn't matter what a or c is equal too. all we need are b and d.

so efe = 1f1. now we need to solve for f. take out 1bb - dd = 1f. find numbers that can equal to 1f while fulfilling the other two. so after a bit of trial and error (sorry i don't have another way), we got 111 - 99 = 12. so now we know that efe is 121.

yay thats the answer

The problem restated: $1000w + 100x + 10x + w - 1000y + 100z + 10z + y = 100q + 10p + q$

where w,x,y,z,q,p are all single digit positive integers, w,y,q non-zero.

Rearranging:

$1001(w-y) + 110(x-z) = 101q + 10p$

Let f = w-y and -g = x-z (f and g must therefore also be single digit positive integers):

$1001f - 110g = 101q - 10p$

Then f must equal q for the ones digit on both sides to be equal:

$900q -110g = 10p$

$90q -11g = p$

Since p must be a single digit and q must be nonzero, 0 < q < 2, so q = 1. Since p must also be a positive single digit, 7< g < 9, so p = 2.

So $C = 101q + 10p = 121$

if A has form a[0]a[1]a[1]a[0] and B has form b[0]b[1]a[1]a[0] where the a[I] and b[j] are digits, then we must have a[0]>b[0] else C would end in a zero (and A>B since C>0). But then C would be 4 digits unless a[1]<b[1], which tells us that a[0] = b[0]+1 (since in this case, when we cancel to perform the subtraction we get rid of the 4th digit). So we now know C has the form 1n1 since we knew a[0]=b[0]+1 which gives the units digit as being 1, and C is a palindrome. Now, we said that a[1]<b[1] (which allowed us to cancel and reduce the number of digits), hence 10+a[1]-b[1]-1=1, since the hundreds digit of C will be formed from subtracting b[1] from a[1] after the latter has had 1 taken off it and 10 added to it (the process of the 2 cancellations necessary to form the 10s and 100s digits). But then a[1]-b[1] = -8, which means the 10s digit n = 10 +-8 = 2, ince we have taken 10 from the hundreds digit of A. This gives 121.

Denote A=abba B=edde and C=xyx (a,b,c,d,e,x,y is a digit 0..9 A>B => a>e if b>d => C have four digits, but C have 3 => b<d

we have: a-e=x 10b-d=y (1) 10b - (d+1)=x =>10b-d = x+1 (2) and a-(e+1)=0=>a-e=1 =>x=1 from (2) => 10b-d=2 and (1) => y= 2

so C=121