Place-value proofs with digits

A two-digit number is not "$tu$" written together — each digit's contribution depends on where it sits. The tens digit is worth ten times its face value, so with tens digit $t$ and units digit $u$:

where: $t$ is the tens digit (a whole number from 1 to 9, since a genuine two-digit number cannot start with 0), and $u$ is the units digit (0 to 9). Reversing the digits does not change what the digits are, only where they sit — the old units digit now occupies the tens place, and vice versa:

Every digit-reversal trick follows from combining these two expressions.

Subtract the reverse from the original:

The right-hand side is $9$ multiplied by the whole number $t-u$, so the difference is always a multiple of $9$ — whatever digits you started with (as long as $t \ne u$, so the answer is not zero).

Add the reverse to the original:

Here the $10t$ and $t$ combine to $11t$, and $10u$ and $u$ combine to $11u$, so the sum is always a multiple of $11$. The same expansion scales up: a three-digit number with hundreds digit $h$, tens digit $t$, units digit $u$ is $100h+10t+u$, used for more advanced digit-swap proofs beyond this section's scope. Notice that both results share the same skeleton: expand, combine, then pull out a common factor with the distributive law. Nothing about the specific digits $t$ or $u$ matters to why the result is a multiple of 9 or 11 — only the place-value gap between the tens and units positions (a factor of 10) does. This is why the trick works for absolutely any two-digit number, not just the ones you happen to try.

Problem. A two-digit number has digits differing by 4. When the number is reversed and the reverse is subtracted from the original, the result is 36. Find all two-digit numbers that satisfy this, then check which one also has a digit sum of 12.

Step 1. Set up the number and its reverse. Let the tens digit be $t$ and the units digit be $u$, so the number is $10t+u$ and the reverse is $10u+t$.

Step 2. Use the subtraction proof. From the core idea, $(10t+u)-(10u+t) = 9(t-u)$. We are told this equals $36$:

Step 3. Solve for the digit difference. Dividing both sides by 9:

This matches the given condition, so it is consistent — but a single equation in two unknowns does not pin down $t$ and $u$ individually; it only fixes their difference.

Step 4. List the possibilities. Since $t$ must be 1 to 9 and $u$ must be 0 to 9 with $t=u+4$, the valid pairs are $(t,u) = (5,1), (6,2), (7,3), (8,4), (9,5)$, giving numbers $51, 62, 73, 84, 95$.

Step 5. Apply the extra clue — digit sum 12 — to pick one. Check each: $5+1=6$, $6+2=8$, $7+3=10$, $8+4=12$ ✓, $9+5=14$. Only $t=8, u=4$ works.

Verify: number $=84$, reverse $=48$; $84-48=36$ ✓ and $8+4=12$ ✓. The number is 84.

Competitive stretch

If instead you were told the sum with the reverse is $132$ and asked to find the digit sum directly, use $(10t+u)+(10u+t)=11(t+u)=132$, so $t+u=12$ — the same digit sum as above, reached from the addition trick instead of the subtraction one, confirming both proofs describe the same number $84$ (since $84+48=132$).

• Writing a two-digit number as the digits placed side by side (like "$tu$" meaning multiplication) instead of the place-value form $10t+u$ — this single mistake breaks every proof that follows.
• Forgetting that $t$ cannot be 0 for a genuine two-digit number, and that both $t$ and $u$ must be whole numbers from 0 to 9 — a computed value like $u=4.5$ or $t=11$ means the given conditions are impossible, not a valid final digit.
• Mixing up which operation goes with which factor: subtracting the reverse gives a multiple of 9, adding the reverse gives a multiple of 11 — both come from the same $10t+u$ setup but combine differently.
• Treating a sum or difference equation (like $t+u=12$) as enough to find a unique number — it only fixes a relationship between the digits, so an extra clue is usually needed, as in the worked example.

Ask the tutor: Got a digit-reversal riddle with a sum or difference clue and want help turning it into the right place-value equation? Ask the AI tutor to walk through it with you step by step.