A digit reversal

We know that $T = 321 \pmod{1000}$ and $S = 100 \pmod{1000}$ . Thus we see that $T^2 - S^2 = 321^2 - 100^2 = 103041 - 0 = 41 \pmod {1000}$

We have $S=12345678910...100\equiv 100\bmod 1000$ $\Rightarrow S^2 \equiv 100^2 \bmod 1000\equiv 0 \bmod 1000$

Also,

$T=1009998...321\equiv321\bmod1000$ $\Rightarrow T^2\equiv 321^2\bmod1000\equiv 41\bmod1000$

$T^2-S^2\equiv 41-0\bmod1000\equiv 41 \bmod1000$

$\therefore Remainder = 41$

We wish to find $T^2 - S^2 \pmod{1000}$ . The last three digits of $S$ are $100$ , so

$S^2 \equiv 0\text{ }(\text{mod } 1000) \implies T^2 - S^2 \equiv T^2\text{ }(\text{mod } 1000).$

The last three digits of $T$ are $321$ , so

$321^2 = 103041 \implies T^2 \equiv \boxed{41}\text{ }(\text{mod } 1000).$

The question is asking for $T^2-S^2 \equiv x \pmod{1000}$ , where x is the remainder. Since it's asking for the last 3 digits, we would only care about the last three digits of those huge numbers.

So $T=321 \text{ and } S=100$ .

Calculating $T^2-S^2 \pmod{1000}$ , we get $\boxed{41}$

Since we are only asked to find the remainder when $T^2 - S^2$ is divided by $1000$ , we need only find the last three digits of each of the numbers; then we can find the answer.

First, since $S$ is the first $100$ numbers written in increasing order without spaces, $S$ has last three digits $100$ . Also, we know that $T$ has last three digits $321$ .

Now we calculate the last three digits of $S^2$ and $T^2$ . Since the last three digits of $S^2$ are only dependent on the last three digits of $S$ , the last three digits of $S^2$ are the last three digits of $100^2 = 10000$ , which is $000$ . Similarly, the last three digits of $T^2$ are the last three digits of $321^2 = 103041$ , which are $041$ .

So, our final answer is $41 - 0 = \boxed{41}$ .

$S = 12345......99100$

$T=10099......54321$

$T^{2} - S^{2} = (T + S)(T - S)$

$= (.....321 + ......100)(.....321 - ......100)$

$= (......421)(.......221)$

$=.........041$

Hence, Remainder = 42

Given that: $S = 12345...99100$ We have: $S ≡ 100 (mod 1000)$ $S^{2} ≡ 100^{2} ≡ 0 (mod 1000)$

Now, given that: $T = 10099..54321$ We have: $T ≡ 321 (mod 1000)$ $T^{2} ≡ 321^{2} ≡ 41 (mod 1000)$

At this point, we have: $T^{2} - S^{2} ≡ 41 - 0 (mod 1000)$ $T^{2} - S^{2} ≡ 41 (mod 1000)$

Thus, our wanted remainder is $41$ .

S = 123....................9899100

T = 1009998...................321

$T^{2}-S^{2}$ = $(T+S)(T-S)$ =(1009998.......321-123......9899100) (1009998.......321+123.....9899100) =(........221)(......421)=..........93041

the number obtained would be exactly divisible by 1000 if there are three 0s at the end so remainder = 41 as when 41 is subtracted from ........93041 the number becomes exactly divisible by 1000

The last two digits of S is "00" , so the last four digits of S^{2} T can be expressed as T=(1000a + 4321) so T^{2} = a\times 10^{6} + 2a\times 4321\times 1000 + 4321^{2} It is obvious that the last three digits of T^{2} is the last three digits of 4321^{2} = 041 Therefore, the answer is 041 ( which is the last three digits of T^{2} - S^{2})

Actually , the question is indirectly asking the last three digits of $T^2 - S^2$ . Finding last three digits of $S^2$ and $T^2$ , $S^2 = \dots0000 , T^2 = \dots041$ . Hence $T^2 - S^2 = \dots041$ and thus , answer is $\fbox{41}$

The rest of the product of two numbers can be of the follo form:

(a * b) % x = ( (a % x) * (b % x) ) % x

and the difference too:

(a - b) % x = ( (a % x) - (b % x) ) % x

where m % x represent the rest of divide the number m between the number x

Since T = ...54321 and S = ...9899100 Then fwe have that T % 1000 = 321 and S % 1000 = 100

then

(T² - S²) % 100 = ((T² % 1000) - (S² % 1000)) % 1000 = (((T % 1000) * (T % 1000)) % 1000 - ((S % 1000) * (S % 1000)) %1000) % 1000 = ((321 * 321) % 1000 - (100 * 100) % 1000) % 1000 = (103041 % 1000 - 10000 % 1000) % 1000 = 41

Last 3 digits of S are 100 so S^2 is divisible by 1000 Last 3 digits of T are 321 . So last 4 digits of T^2 are 3041 . when divided by 1000 remainder is 41

Note that any number is congruent to its last three digits modulo $1000$ , hence $$S \equiv 100 \pmod{1000}$$ $$T \equiv 321 \pmod{1000}$$ so $$T^2 - S^2 \equiv 321^2 - 100^2 \equiv 221 \cdot 421 \equiv \boxed{41} \pmod{1000}$$

Since we need only $T^2 - S^2$ mod 1000 we can just look at $T$ and $S$ each mod 1000.

The last three digits of $T$ are $321$ , so $T^2 = 041$ mod 1000. Similarly, $S^2 = 000$ mod 1000, so the desired answer is $041 - 000 = 041$ .

First of all, let us notice that $S=1234\dots 9899100$ and $T=1009998\dots 4321$ . Now, in order for a number to be divisible by $1000$ we know that the last $3$ digits of the number need to be zeros, which we will use later on. Also, since $T$ and $S$ are very big integers, we will only use the last couple of digits in them to solve the problem, since the rest are not needed at all.

Therefore

$S^{2}=\overline {abcd\dots 3610000}$

$T^{2}=\overline {mfgn\dots 9971041}$

On the other hand, $T^{2}-S^{2}=\overline{pqrs\dots 6361041}$ .

According to the Remainder Theorem, $x=yq+r \Rightarrow x-r=yq$ , where $r$ is the remainder. This means that if a number divided by another one has a remainder, then that number minus the remainder is divisible by the original divisor. If we apply the theorem to our problem, we'd get

$\overline{pqrs\dots 6361041} - r=1000q$

Since $T^{2}-S^{2} - r$ needs to be divisible by $1000$ , we apply the divisibility rule mentioned above. Consequently, the only way to get the last $3$ digits of $T^{2}-S^{2}$ to be zeros is to have the remainder be $41$ (since $\overline{pqrs\dots 6361041} - 41 = \overline{pqrs\dots 6361000}$ ), meaning that finally $r=41$ .

last 3 digits of S^2 are 0000 and last 4 digit of T^2 are 1041 so when you subtract S^2 from T^2 , the last 3 digits will be 041..