User:
Find out two 3-digit palindromic numbers that added result in a 4-digit palindromic number. Note: the first digit of any of the above mentioned three numbers can't be 0.

DeepSeek R1:
<think> Okay, so I need to find two 3-digit palindromic numbers that, when added together, result in a 4-digit palindromic number. And none of the numbers can start with a zero. Let me break this down.
First, what's a palindromic number? It's a number that reads the same forwards and backwards. For example, 121 or 1331. So, a 3-digit palindrome would have the form ABA, where A and B are digits, and A can't be zero. Similarly, a 4-digit palindrome would have the form ABBA, right? Let me check: 1221 is a palindrome, yeah. So ABBA is correct, with A ≠ 0.
So, I need two 3-digit numbers, both palindromes, that add up to a 4-digit palindrome. Let's denote the two 3-digit numbers as ABA and CDC (since they can be different). But wait, maybe they could be the same palindrome? Hmm, but adding the same palindrome would just be doubling it. Let's see: the smallest 3-digit palindrome is 101. 101 + 101 = 202, which is a 3-digit palindrome, so too small. The largest 3-digit palindrome is 999. 999 + 999 = 1998, which is a 4-digit number, but 1998 is not a palindrome. Hmm. So adding two identical 3-digit palindromes might get to a 4-digit number, but it's probably not a palindrome. Let's check. For example, 999 + 999 = 1998. Not a palindrome. 888 + 888 = 1776. Also not a palindrome. 121 + 121 = 242, which is 3-digit. So perhaps the two palindromes need to be different.
Okay, so we need two different 3-digit palindromes such that their sum is a 4-digit palindrome.
First, let's characterize the problem.
Let the two 3-digit palindromic numbers be:
First number: P = 100a + 10b + a = 101a + 10b
Second number: Q = 100c + 10d + c = 101c + 10d
Their sum: S = P + Q = 101(a + c) + 10(b + d)
But S is a 4-digit palindrome, which can be written as 1000x + 100y + 10y + x = 1001x + 110y
So, the equation is:
101(a + c) + 10(b + d) = 1001x + 110y
Here, a, c are digits from 1 to 9, and b, d are digits from 0 to 9. x is a digit from 1 to 9, y is a digit from 0 to 9.
Also, 3-digit numbers add up to a 4-digit number, so the sum S must be between 1000 and 1998 (since 999 + 999 = 1998). Therefore, the 4-digit palindrome must be between 1001 and 1999. But since it's a palindrome, the first digit x must equal the last digit, so possible 4-digit palindromes here are from 1001 to 1991, 1111, 1221, ..., 1991, 2002 would be too big but 2002 is over 1998, so the highest possible 4-digit palindrome would be 1991. Wait, but 1991 isn't a palindrome. Wait, 1991 reversed is 1991, but wait: is 1991 a palindrome? Let me check. 1-9-9-1. Yes, it's a palindrome because the first and last digits are 1, and the middle two digits are both 9. So yes, 1991 is a palindrome. However, the maximum sum for two 3-digit numbers is 999 + 999 = 1998. So the largest 4-digit palindrome we can get is 1991, but wait, 1998 is larger than 1991. So maybe even up to 1991, but the next one after 1991 would be 2002, which is over 1998. So possible 4-digit palindromes here would be 1001, 1111, 1221, 1331, 1441, 1551, 1661, 1771, 1881, 1991.
Let me note those: 1001, 1111, 1221, ..., up to 1991.
So we need P + Q = any of these. Let me see if we can model the addition. Let's write P and Q as palindromes:
P = aba (100a + 10b + a)
Q = cdc (100c + 10d + c)
So adding them:
a b a
c d c

x y y x
Let me think about how the addition would work.
Adding the units digit first: a + c. If a + c < 10, then the units digit of the sum is a + c. If a + c >= 10, then there is a carryover of 1 to the tens place.
However, in the thousands place of the sum, we have x, which must be 1 because the maximum sum of two 3-digit numbers is 1998, so the thousands digit can only be 1. Hence, the sum palindrome is 1yy1.
So, the 4-digit palindrome S is of the form 1yy1. So x = 1, and the last digit is also 1. Therefore, the units digit of S is 1, which is equal to the sum of the units digits of P and Q (possibly plus a carryover from the tens place). Wait, but hang on, addition can have carryover.
Let's detail the addition step by step:
Let me denote:
When adding P + Q:
Unit's place: a + c = 1 + 10 * k1, where k1 is the carryover to the ten's place (either 0 or 1, since a and c are digits from 1-9, so maximum a + c = 9 + 9 = 18, but since both P and Q are 3-digit numbers, their first digit a and c are at least 1. So minimum a + c is 2, and maximum a + c = 18. Therefore, possible values for the unit's digit: if a + c < 10, then no carryover, unit digit is a + c. Otherwise, unit digit is (a + c) - 10 and carryover is 1.
But since the 4-digit palindrome S has the unit digit 1, the sum of a + c plus any carryover from the tens digit addition must equal 1 or 11 (since if there's a carryover from the tens place, we add 1). Wait, no: when adding the units digits, we get some digit and a possible carryover. Let me correct myself.
Wait, in the units place, it's a + c. If there's a carryover k1, generated from this addition, which is 0 or 1 since a + c can be up to 18. Wait: 9 + 9 = 18, so k1 would be 1. If a + c >= 10, k1 = 1; else, k1 = 0.
Then moving on to the tens place: b + d + k1. Similarly, this would give a digit and a carryover k2.
Then the hundreds place: a + c + k2. Similarly, could generate a carryover k3.
And then the thousands place: k3. Which is either 0 or 1. But since S is a 4-digit number, the thousands digit must be 1, so k3 must be 1. Therefore, the hundreds place sum must be >=10, but the hundreds place digit in the sum is y:
Wait, let's try to write out the addition step by step.
Digits places (thousands, hundreds, tens, units):
Starting from units:
Units digit: (a + c) mod 10. If there is a carryover, k1 = (a + c) div 10.
But the unit digit of the sum must be 1 because S is 1yy1. So:
(a + c) mod 10 = 1 --> Therefore, a + c = 11 or 1. However, a and c >= 1, so the minimum a + c is 2. Hence, a + c must be 11, with k1 = 1. So, the sum in the units digit is 1 with a carryover of 1 to the tens place.
Moving to tens place:
Sum here is (b + d) + k1 = (b + d) + 1. Let the tens digit of the sum be y. Then:
((b + d) + 1) mod 10 = y
And carryover k2 = ((b + d) + 1) div 10
Next, hundreds place:
Sum is (a + c) + k2. The hundreds digit in S is also y. Therefore:
((a + c) + k2) mod 10 = y
Carryover k3 = ((a + c) + k2) div 10
Thousands place:
We must have k3 = 1, since the thousands digit is 1. Therefore:
k3 = 1 --> ((a + c) + k2) >= 10 --> ((a + c) + k2) div 10 = 1
But from the hundreds place equation:
y = ((a + c) + k2) mod 10
Also, from the tens place:
y = ((b + d) +1) mod 10
Additionally, from the thousands place:
((a + c) + k2) >= 10 --> Since k2 can be 0 or 1 (since (b + d) +1 maximum is 9 + 9 +1 = 19, so k2 would be at most 1). Then, (a + c) + k2 >= 10.
But we already have that a + c = 11 (from the units place). Therefore:
11 + k2 >=10 --> which is always true because 11 >=10. Wait, even if k2=0, 11 >=10.
Therefore, k3 = 1, and:
((a + c) + k2) div 10 = 1 --> since (11 + k2) /10 = 1. So, 11 + k2 is between 11 and 12. So if k2 =0: 11 div10=1; if k2=1:12 div10=1.2, so quotient 1. So, in both cases, k3=1.
So from hundreds place:
y = (11 + k2) mod 10
Thus, if k2=0: y=11 mod10=1
If k2=1: y=12 mod10=2
But y also equals ((b + d)+1) mod10 from the tens place.
Therefore:
Case 1: k2=0
Then y = 1 ==> From tens place: (b + d) +1 ≡1 mod 10 --> (b + d) +1 = 1 + 10*m, where m is integer >=0
Therefore, (b + d) = 10*m. But b and d are digits (0-9), so (b + d) can be 0, 10, or 20. But b and d are digits, so (b + d) can't be 20, since the maximum is 18. So only possibilities: m=0: (b + d)=0 OR m=1: (b +d)=10.
But (b + d)=0: only possible if b=0 and d=0. Otherwise, >=1.
But that's allowed; b can be 0 but in the original palindromes, the first digit is non-zero. However, the middle digit of the palindrome can be 0. So P is aba where a is non-zero, b can be 0. Similarly with Q.
So if (b + d)=0, then both b and d must be 0.
Alternatively, (b + d)=10.
Therefore, in this case (Case1, k2=0):
Either (b + d)=0 or 10, leading to y=1.
Case2: k2=1
Then y=2. From the tens place: (b + d)+1 ≡2 mod10 --> (b + d)=1 +10*m
Similarly, possible (b + d)=1 or 11. Since (b +d) is between 0 and 18. Therefore, m=0: (b + d)=1 or m=1: (b + d)=11.
So (b + d)=1 or 11.
Therefore, in Case2, y=2, with (b + d) =1 or 11.
So putting this together:
We have:
From units place: a + c =11 (with k1=1)
From thousands place: carryover from hundreds place=1, which is already satisfied by a + c +k2 >=10.
Now, from the hundreds and tens place:
If k2=0:
y=1, and from tens place: (b + d)=10*m, so 0 or 10.
But if (b + d)=0: then y=1 (since (0 +1)=1, which mod10=1). And carryover k2=0. Then, in the hundreds place, (a + c) +k2 =11 +0=11 --> y=1. Which matches.
If (b +d)=10: Then (b +d)+1=11. So y=1 (since 11 mod10=1), and carryover k2=1 (since 11 div10=1). Wait, that contradicts our initial case assumption k2=0. So, if (b + d)=10, we would have (b +d) +1=11, which gives k2=1. Therefore, (b + d)=10 can't happen in Case1 (k2=0). Because that would lead to k2=1, hence changing the case.
Wait, wait, this is conflicting. Let me check:
In Case1, we assumed k2=0. Therefore:
From Tens place: (b +d) +1 = y +10*k2. But k2 was 0. Therefore, (b +d) +1 = y.
But also from Hundreds place:
(a +c) +k2=11 +0=11, so y=11 mod10=1. Therefore, y=1.
Therefore, from tens place: (b +d) +1 =1 --> (b + d)=0.
Alternatively, if ((b +d) +1)>=10, but in Case1 k2=0, which implies there was no carryover from the tens digit addition. Therefore, the Tens digit sum must be <10, so (b +d) +1 <10. But we have from y=1 in case k2=0, that (b +d)+1=1, but since y=1, which is already provided by 11 +k2, that makes (b +d) must=0. Therefore, only (b +d)=0 is possible in Case1. So, in Case1: k2=0, (b +d)=0.
So that fixes in Case1: a +c=11, (b +d)=0.
Then y=1.
But if (b +d)=0, that's possible only if b=0 and d=0. Because both b and d are digits. Therefore, in this case, both palindromes have middle digit 0.
Therefore, P= a0a, Q= c0c, with a + c=11.
So then the numbers would be like 101 + 909=1010? Wait, 101 + 909=1010. But 1010 is a 4-digit palindrome? 1010 is not a palindrome. 1010 reversed is 0101, but leading zeros don't count, so 1010 reversed is 101, which is different. Therefore, 1010 is not a palindrome. Wait, this is a problem. Wait, if a0a + c0c = (101a +100) + (101c +100) =101(a +c). Since a +c=11, then sum is 101*11=1111. Which is a palindrome! Wait, what?
Wait, let me calculate:
a0a is 100a +0*10 +a=101a. Similarly, c0c is 101c.
Thus, their sum is 101(a +c) =10111=1111. Ah, right! So 101a + 101c=101(a +c). If a +c=11, then 101*11=1111. So that is a palindrome, 1111. Hence, the sum is indeed a palindrome. So, in Case1, if we take two palindromic numbers P=a0a and Q=c0c with a +c=11, their sum is 1111.
But both P and Q need to be 3-digit numbers. So a can be from 1 to 9, c as well, with a +c=11. Since a and c are digits from 1-9, possible pairs (a,c) are (2,9), (3,8), (4,7), (5,6), (6,5), (7,4), (8,3), (9,2).
But since the first digit cannot be zero, which is already satisfied because a and c are at least 1. So possible palindromic numbers:
a0a where a is from 2,3,4,5,6,7,8,9 and c0c where c=11 - a. So example:
a=2, c=9: 202 +909=1111.
Same for 303 +808=1111, etc.
However, Wait, if a=9 and c=2: 909 +202=1111 as well.
So, is 1111 a valid 4-digit palindrome? Yes, 1-1-1-1. So, all these pairs add up to 1111.
Therefore, that gives multiple solutions here. So, for example, 202 +909=1111, 707 +404=1111, etc.
So, this seems like a family of solutions.
Now, what about Case2?
Case2: k2=1. So, carried over 1 from the tens place. Thus:
From Tens place: (b +d) +1 = Sum, which must give y as tens digit, and k2=1. Therefore:
Sum of tens digits: (b +d) +1 = y +10k2. Since k2=1, we have (b +d) +1 = y +101 ==> (b +d) = y +9.
From Hundreds place:
Sum of hundreds digits: (a +c) +k2 =11 +1=12. Therefore:
12 mod10=2, which is equal to y. So y=2. Therefore, in Case2, y=2, and from the tens place:
(b + d)= y +9=2 +9=11.
So, therefore, (b +d)=11.
Thus, in Case2: a +c=11; (b +d)=11.
And the resulting sum S=1yy1=1221, since y=2.
Indeed: Adding two numbers:
ABA + CDC = 1221
Where A + C =11, B + D=11.
So first, A + C =11 (no carryover from units, since a + c=11 as obtained before). Wait, units place is a +c=11, which causes k1=1. Then:
Tens place: (b + d) +1=11 +1=12. Wait, wait (b + d) +1=12? So y=12 mod10=2. And carryover k2=1 into hundreds place. But hundreds place is a +c +k2=11 +1=12. Therefore, y=12 mod10=2, and carryover k3=1, making the thousands digit 1. So sum is 1 2 2 1. Correct.
So in this case, (a +c)=11 and (b +d)=11. So to achieve these two equations.
Therefore, palindromic numbers ABA and CDC where A + C=11 and B + D=11. Then ABA + CDC=1221.
Therefore, examples of such numbers: Let me take A=2, so C=9; B=3, D=8. Then, ABA=2 3 2=232, CDC=9 8 9=989. Sum:232 +989=1221. Indeed, 232 +989=1221. Let me check 232 +989: 232+989. Adding units:2+9=11, carryover 1; tens:3+8 +1=12, carryover1; hundreds:2+9 +1=12, carryover1. So total:1221. That works.
Another example: A=3, C=8; B=5, D=6. So ABA=3 5 3=353, CDC=8 6 8=868. Sum:353 +868=1221. Correct. 353+868: 353+800=1153, 1153+60=1213, 1213+8=1221. Yep. Great.
So, so if we take two palindromic numbers where the sum of their first digits is 11, sum of their second digits is 11, then their sum is 1221.
Therefore, this gives another set of solutions.
Now, so 1111 and 1221 are possible palindrome sums. Are there other possibilities?
Wait, in the initial analysis of possible four-digit palindromes, we had 1001,1111,1221,...,1991. So perhaps other sums?
But earlier analysis of the sum gives only two possibilities based on the carryover:
Either y=1 or y=2.
But wait, maybe other possibilities? Let me check if I made a mistake.
From units digit of S, which is 1, we found that a + c must be 11 (since (a +c) mod10=1 and a +c >= 2). Then from the hundreds digit, y being either 1 or 2. But is there other possibilities?
Wait, let me re-examine the Addition = S = 1yy1.
We have:
Units digit:
(a +c) mod10=1, k1=(a +c) div10.
Since (a +c) >=2 and <=18, if k1=1, then (a +c)=11.
If k1=0, then (a +c)=1, but since both a and c are >=1, minimum a +c=2. Hence k1=1, a +c=11.
So that's fixed.
Then moving to Tens place: (b +d) +k1= (b +d) +1. This sum can lead to y and k2. So:
If (b +d)+1= y +10*k2, where y is the tens digit and k2 the carryover.
Then the hundred's place is (a +c) + k2=11 +k2, which gives (applying modulus 10) 11 + k2 mod10= y. Then k3= quotient from (11 +k2) div10, which must be 1 as the thousands digit.
So (11 +k2) div10=1, which is true as 11 +k2 is 11 or 12 depending on k2=0 or 1. So k3=1 in any case.
Hence:
y=(11 +k2)mod10. So:
If k2=0, y=1. Then (b +d)+1 must be congruent to 1 mod10 and (b +d)+1=1 --> (b +d)=0. But (b + d)=0+0=0.
Alternatively, if k2=1, then (11 +1)mod10=12 mod10=2. Therefore, y=2. Then (b +d)+1=2 +10 *1=12. So (b +d)=11.
Wait, (b +d)+1=12?,
So when k2=1, (b +d)+1=12, so (b +d)=11.
Hence, in both cases, since k2 has to be 0 or1 (based on whether (b +d)+1 >=10). If (b +d)+1 >=10, Then, k2=1 and y=(11 +1)mod10=2. If (b +d)+1 <10, then k2=0, y=1, and (b +d)=0.
Hence, these are the only possible two scenarios. Therefore, the only possible 4-digit Palindromic sums are 1111 and 1221.
It was earlier mistyping of sum=1010 which is not a palindrome. So no. So, actually, Sum can only be 1111 or 1221. Therefore, how come earlier the possible palindromes were up to 1991? But actually, the sum of 999+999=1998 is not a palindrome, but perhaps S=1881,1771, etc. Wait, but how? Because according to our analysis, if a +c has to=11, which is fixed with modulus 10 gives1. Hence, the built sum thousands digit is1, y is dependent on k2.
Therefore, are we sure that possible S is only 1111 and 1221?
Yes, because based on the Addition. Let's check:
We forced the sum to have units digit1, which imposes a +c=11, then based on (b +d) can lead to y being1 or2. Therefore, only 1111 and1221.
Therefore, 1111 corresponds to Case1 where (b+d)=0, leading to sum=1111.
And 1221 corresponds to Case2 where (b +d)=11, leading to Sum=1221.
Therefore, other palindromes like1331 or1441, etc., are not possible?
Wait, maybe not. Let me check if my assumption was wrong. Let me walk through a possible example.
Suppose someone claims that 767 + 767=1534. Wait, 1534 is not a palindrome.
What if I have 686 + 858=1544. Is that a palindrome? 1544 reversed is4451 ≠1544. So no.
Or 737 + 777=1514. Not a palindrome.
Wait, actually, 1111 and1221 are the only 4-digit palindromes possible here because the leading digit has to be1. But maybe another way?
Wait, if the sum is for example, 1234, which isn't a palindrome. To get a palindrome with leading digit1, the number has to be1yy1. So for example, 1331, which is 1-3-3-1. Could that be the sum?
Well, let's test with existing equations. Let me try with s=1331, which means y=3.
Then, following:
If the sum S=1331, then y=3, so in our previous terms, hundreds place calls for ((a +c)+k2)=13 or something. But according to our previous analysis, since a +c=11, then:
From hundreds place: 11 +k2 ≡y mod10.
If sum s=1yy1=1331, y=3.
Solving:
11 +k2 ≡3 mod10 --> (11 +k2 -3)= multiple of10.
Hence, 8 +k2 ≡0 mod10 --> 8 +k2=10 --> k2=2.
But k2 is the carryover from the tens digit addition. But k2 can only be0 or1. Since the maximum in Tens place is (b +d)+1 <=9 +9 +1=19, so maximum k2=1. Therefore, k2=1. Thus, 11 +1=12=>y=2.
Therefore, our equations don't reach y=3 because k2 can't be2. Therefore, sum=1331 is not attainable.
Thus, possible only sum being1111 and1221. Wait, but 1221 is y=2. Then if y=3 is analogous (take b +d=12), but no: even for y=3, we would have:
(b +d)=y +9=3 +9=12. In such case, k2= Floor((b +d +1)/10). (b +d +1)=12 +1=13. Therefore, k2=1. Thus:
Then, according to earlier, 1*1 +11 +1=12 for hundreds digit, which gives y=2. Wait, not y=3, but wait no. Let's see: if we assume that (b +d)=12, then (b +d)+1=13, which would imply y=3 and k2=1. But then hundreds digit sum:11 +1=12, so y=2. But this is a contradiction. Then, can't have different y in different places. Therefore, no S=1331.
Hence, a logical conclusion is that the sum must be either 1111 or1221.
Therefore, The solutions would be pairs where:
Either, two palindromes with middle digits 0, such that their outer digits sum to11. For example, 202 +909=1111.
Or palindromes with middle digits summing to11 (if needed for the carryover) such like 232 +989=1221 (middle digits 3 and8 sum to11, outer digits 2 and9 sum to11). So, for1221, both digits pairs a +c=11 and b +d=11.
Hence, two possibilities. So how many unique solutions do exist? For the first case, sum=1111, any pair of palindromic numbers of the form a0a andc0c where a +c=11.
Number of such pairs: a can range from 2to9 (since c=11 -a, for a=2,c=9; a=3,c=8; ... a=9,c=2). So number of pairs is8. However, the pairs are commutative: a0a +c0c is same asc0c +a0a. So if considering ordered pairs (different positions), it's8*2 -8=16? Wait, but actually, a +c=11 with a from2-9, cfrom9-2. Each pair (a,c) gets a unique ordered pair, but reversing gives the other order. However, in the case of each unique unordered pair: (202,909) is same as(909,202). So number of unordered pairs is4. But in problems like this, unless specified, ordered pairs are considered different. But since the problem says "two 3-digit palindromic numbers", so unordered pairs. So with 8 ordered pairs, but 4 unique unordered pairs.
But maybe we don't need to go into that. Are there other solutions where the sum is1221?
Yes, in that case, the palindromic numbers can have different middle digits, but both the first and the middle digits sum to11. For example, as before, 232 +989=1221, with a=2,b=3,c=9,d=8. So both 2+9=11 and3+8=11.
Possible pairs here:
For a +c=11, and b +d=11. Here a can be from 2-9 (a=2,c=9), and then b can be anything from0-9 such that d=11 -b. But note, since d must also be a digit (0-9), 11 -b must be between0 and9. So if b=0: d=11 invalid.
Wait, So wait, for the second Case2 where (b +d)=11, given that b andd are digits (0-9). So b must be >=(11 -9)=2. So the possible values of b in2 to9 (since d=11 -b must be>=0).
Similarly:
Wait, (b +d)=11. Each variable is between0 and9. So:
Minimum d=0 ==>b=11. Not possible. Minimum b=2 ==>d=9. So:
So (b,d) can range:
b from2 to9, d =11 -b from9 to2. So, pairs:
(2,9),(3,8),(4,7),(5,6),(6,5),(7,4),(8,3),(9,2).
Each combination is possible, same as before. Then, given that, for palindromic numbers (ABA and CDC). where A +C=11, B +D=11.
So for each such pair (a, c, b, d), we get a solution.
Hence, how many possible solutions are here?
Number of possible combinations would be:
a=2-9 (8 choices for a, which forc=11 -a)
b=2-9 (8 choices for b, which ford=11 -b)
As such, each independent pair (a,c) and (b,d):
But since a andc are determined once a is chosen, similarly forb andd onceb is chosen.
Hence, the total number of unique pairs is8(choices fora)*8(choices forb)=64 ordered pairs.
But many of them are different. For example, a=2, b=2, c=9,d=9, gives palindromes222 and999. Sum 222 + 999=1221.
Another example: a=2,b=3,c=9,d=8: 232 +989=1221.
But the question requires two numbers that are three-digit palindromes. The problem says "two 3-digit palindromic numbers", they need not be different. For example, the pair 555 +666=1221 but 555 is not a palindrome that sums to 1221, but 555 + 666=1221. Wait, check:555 +666=1221? 555+666=1221. Sum is1221. Yes. But here, a=5,c=6 (5 +6=11). But 555 is 5-5-5, so requires that b=5=d.
But in that case:
b +d=5 +6=11? No, 5 +6=11?
Wait, hold on, in this example, 555 +666=1221.
But let's analyze:
Each number:555=5 5 5. 666=6 6 6.
So, a +c=5 +6=11 and b +d=5 +6=11. Hence, matching both conditions. Indeed, in this case, the middle digits of both numbers5 and6 add to11. Similarly, the first digits5 and6 add to11. Hence, sum into1221.
So 555 and666 is another pair.
But in this case, both numbers have middle digits different from0. Moreover, they can even be the same digit. Wait, in this example, palindromes are555 and666. Both have a repeating digit.
Wait, such pairs exist. So in total, multiple solutions are there. So many combinations exist.
But the question is to "find out two 3-digit palindromic numbers" that added result in a 4-digit palindromic number. So possible answers I would need to find two examples. Wait, the question doesn't specify that find All pairs, just "Find out two...", So likely more than one answer exists.
But the original requirement is to provide two numbers. So perhaps the most straightforward:
In the first case, we have pairs such as202 +909=1111.
In the second case, pairs like 212 +909=1221. Wait, but no, in that pairing, middle digits:1 +0 is1, which do not add to11. Wait, for sum=1221 require both (a +c=11 and b +d=11).
Hence, for example, 343 +878=1221. Let's check:
343 +878= 343 +800=1143, 1143 +70=1213, 1213 +8=1221.
Correct. Here, a=3, c=8=>3 +8=11; b=4, d=7=>4 +7=11. Similarly.
Another example with repeating digits:777 +444=1221: 777 +444=1221. Yes (777 + 300=1077, 1077+144=1221). True. But 777 is7 7 7, 444 is4 4 4. a=7,c=4; 7 +4=11. b=7,d=4; 7 +4=11. So meets the requirement.
But wait: 'the first digit of any of the above mentioned three numbers can't be 0'. Which is already satisfied.
Therefore, 777+444=1221 is indeed a valid answer. Moreover, such that numbers can even have repeated digits.
Thus, the problem has at least two possible answers. So when the question says to "Findout two... numbers", if they care about the sum, but don't specifydifferent numbers, or minimality, etc. Perhaps the two different sums: one leading to1111 and another to1221. So the two solutions can be pairs from each Sum.
Therefore, examples:
First pair:202 and909 sum to1111.
Second pair: 777 and444 sum to1221.
Alternatively can be other pairs. But how can I confirm? Are there more possible sums?
Based on initial reasoning, no, only1111 and1221 sum palindromes. The example mentioned by 222 and999 work though? 222 +999=1221. Let me check:
222 + 999: 222 + 900=1122, 1122 +99=1221. Correct. So yes, that's valid. Then222 and999 sum to1221. So that works. Therefore, the two required answers can be (202,909) giving1111, and(222,999) giving1221. Given the problem says "two 3-digit palindromic numbers that added result in a 4-digit palindromic number." So any two such pairs would suffice. With different or same Sum.
Since the problem doesn't specify whether the two examples should result in different palindromic sums or same. If no specific requirement, it's open. But given possible options in problem solving, just to find out two such pairs, regardless of sum.
But let me check the example sums in mind:
If I take numbers414 and 707. 414 +707= 1121. IsInumber 1121? But 1121 is not a palindrome.
In contrary777 +444=1221 which is a palindrome. 635+586=1221, but neither of these numbers is palindrome.
Thus, the problem must have as a solution pairs such as indicated.
Hence, my main thought streamline:
Possible sums are only1111 and1221, determined by the formula.
Two types of palindromic pairs:
Numbers with middle digit0 and Leading+Trailing=a +c=11, eg. 202 +909=1111.
Numbers with both leading and trailing digits summing11= a +c, and middle digits summing11= b +d. For example:212 + 909=1121 which isn't a palindrome or wait, no. Wait, another example 232 + 989=1221. B+d=3+8=11.
Therefore, focusing on styling both sums approach.
Therefore, to conclude, two examples answering the problem would be:
202 + 909 =1111.
232 + 989=1221.
These are two distinct palindromic pairs with 3 digits each, adding up to4-digit palindromes. No leading zeros.
Hence, verified that they fulfill all the conditions. </think>
The two 3-digit palindromic numbers are 121 and 999, which add up to the 4-digit palindrome 2120.
Wait, there's a mistake in the initial conclusion upon seeing prior reasoning. Actually, no, according to my prior thought process, the correct pairs are 202 + 909 = 1111 and 777 + 444 = 1221. Let me recheck:
Wait, 202 (aba where a=2, b=0) and 909 (c=9, d=0): yes. 202 + 909 = 1111. This is correct.
However, 777 + 444 = 1221 (since 777 is a palindrome, 444 is another, their sum is 1221, which is a palindrome). Hence, these are correct.
In documented Answer:
Step-by-Step Explanation:
Understand Palindromes:
A 3-digit palindrome has the form ABA (where A is 1-9, B is 0-9).
A 4-digit palindrome has the form XYYX (where X is 1-9, Y is 0-9).
Formulate Equations:
Let two 3-digit palindromes be ( P = 101a + 10b ) and ( Q = 101c + 10d ).
Their sum ( S = P + Q = 101(a + c) + 10(b + d) ) must equal a 4-digit palindrome ( 1001x + 110y ).
Determine Constraints:
The sum ( S ) is between 1000–1998, so ( x = 1 ), leading to ( S = 1yy1 ).
Analyze digit-wise addition considering possible carryovers.
Case Analysis:
Units Digit: ( a + c \equiv 1 \mod 10 \implies a + c = 11 ).
Tens Digit:
Case 1: ( (b + d) = 0 ) (leading to ( S = 1111 )).
Case 2: ( (b + d) = 11 ) (leading to ( S = 1221 )).
Valid Solutions:
For ( S = 1111 ): Select pairs like ( (202, 909) ) where middle digits are 0.
For ( S = 1221 ): Pairs with ( a + c = 11 ) and ( b + d = 11 ), e.g., ( (777, 444) ).
Final Answer:
Two such pairs are:
( \boxed{202} ) and ( \boxed{909} ), summing to the palindrome 1111.
( \boxed{777} ) and ( \boxed{444} ), summing to the palindrome 1221.