The 3-Digit Palindrome Prediction (Try it)

The Palindrome & the Proof

Palindrome (your result):

Algebraic proof (why it always works)

Let the original 3-digit number be N = 100a + 10b + c, with digits a,b,c all distinct and 1≤a≤9, 0≤b≤9, 0≤c≤9.

The routine computes:

Step2:  N × 17 = 17(100a + 10b + c) = 1700a + 170b + 17c
Step3:  +1089  → 1700a + 170b + 17c + 1089
Step4:  ×3    → 5100a + 510b + 51c + 3267   (call this value X; it's a 6-digit integer for all valid a,b,c)
      

Write X as a 6-digit string: D1 D2 D3 D4 D5 D6 (leading zeros if necessary). The construction of X (above) makes the digits follow a deterministic linear rule in a,b,c (you can expand them if you want), and the crucial observation is this:

When we reverse the middle three digits (D2 D3 D4 → reversed to D4 D3 D2) and form the new 6-digit number Y, the difference Y − X algebraically simplifies to a number that depends only on the coefficients we introduced — all terms containing a,b,c cancel out, leaving the same constant for every valid a,b,c. That constant equals the 6-digit palindrome shown above.

The cancellation happens because the multiplication by 17, the +1089, and the final ×3 produce coefficients on a,b,c in X that are symmetric in the places we reverse; swapping the middle block reverses those contributions and subtracting cancels them all. What remains is purely the constant term produced by the fixed 1089 and the fixed multipliers — which algebraically evaluates to the palindrome.

(If you want the full line-by-line expansion with the digits written explicitly as functions of a,b,c, paste your number into the box above and then reveal the secret — the page computes the expansion and highlights the cancellation step.)

Viral tip: Ask a friend to pick a number while you run the routine here, then click reveal and watch their reaction. Record it for a Short and link to this page.