Place Value & Digit Problems

Let the digits be named from the left.

• A two-digit number is $10a+b$, with $1 \le a \le 9$ and $0 \le b \le 9$.
• A three-digit number is $100a+10b+c$.
• The digit sum is $a+b$ (or $a+b+c$), and the digit product is $ab$.

Two constraints are always in force and are the source of most eliminations: the leading digit cannot be $0$, and every digit is an integer from $0$ to $9$.

For $N = 10a+b$ with reversal $N' = 10b+a$:

• $N + N' = 11(a+b)$ — always a multiple of 11;
• $N - N' = 9(a-b)$ — always a multiple of 9;
• so $N=N'$ exactly when $a=b$, and the difference is 0, 9, 18, ... only.

Given a sum and a difference you recover the digits immediately, because you then know $a+b$ and $a-b$.

For $N = 100X+10Y+Z$:

• full reversal: $N - \overline{ZYX} = 99(X-Z)$, so the difference is a multiple of 99 and the middle digit is irrelevant;
• swapping the first two digits: $N - \overline{YXZ} = 90(X-Y)$;
• swapping the last two digits: $N - \overline{XZY} = 9(Y-Z)$.

Every multiple of 99 below 1000 — namely 99, 198, ..., 891 — has middle digit $9$ and outer digits summing to $9$, which is a fact CDS has used directly.

For $\overline{XYZ} + \overline{YZX} + \overline{ZXY}$:

Repeating a block is multiplication by a repunit-style constant:

• $\overline{XYXYXY} = \overline{XY} \times 10101$, and $10101 = 3 \times 7 \times 13 \times 37$;
• $\overline{abcabc} = \overline{abc} \times 1001$, and $1001 = 7 \times 11 \times 13$;
• $\overline{abab} = \overline{ab} \times 101$, and $101$ is prime.

For a block that is repeated but not cleanly — $\overline{XXYXX} = 11011X + 100Y$, where $11011 = 7 \times 11^2 \times 13$ — expand rather than guess, then use divisibility on the pieces.

Write $R_n$ for the repunit with $n$ ones:

• A string of $n$ copies of the digit $d$ is $d \times R_n$.
• The digit sum of $10^n-1$ is $9n$, because it is $n$ nines.
• $R_{2n} = R_n \times (10^n+1)$, which is the identity behind the $x-y^2$ question.

Working modulo $10^k$ keeps exactly the last $k$ digits, and modular arithmetic respects addition and multiplication. So:

• to get the last $k$ digits of a product, reduce each factor mod $10^k$ first, multiply, then reduce again;
• the same holds for sums and for powers.

A useful special case: for $n \ge 3$, powers of 5 settle into a pattern on their last three digits ($5^4 = 625$, and every even power from the fourth onwards ends in $625$), which is why the hundreds digit of $25^{10}$ can be read off without computing the number.

Procedure:

• Expand every numeral into place-value form so the unknown digits appear as ordinary variables.
• Collect terms. A column sum usually produces something like $10(P+Q+R) + \text{constant}$.
• Apply the digit bounds $0 \le \text{digit} \le 9$ and any leading-digit restriction, then test the survivors.

For a product with a near-round multiplier, use the complement trick: $999 \times n = 1000n - n$, which makes the tail of the product easy to control.