The Torn Ledger

Question

An auditor discovers that 25 leaves (single physical sheets) have been ripped out of a paper trade ledger. Each leaf carries two page numbers, front and back, and the ledger is paginated normally: page 1 is the front of the first leaf, page 2 its back, page 3 the front of the second leaf, and so on. The clerk under investigation swears that the 50 missing page numbers add up to exactly 2026. Can t

Accepted Answer

No. The clerk is lying. The invariant. The k-th leaf carries pages 2k - 1 (front, odd) and 2k (back, even), so a single leaf's two pages sum to (2k - 1) + 2k = 4k - 1, which is odd, whatever k is. The 25 missing leaves therefore contribute a sum of 25 odd numbers, and a sum of an odd count of odd numbers is odd. The claimed total 2026 is even. No selection of 25 leaves can produce it. ∎ A sharper screen, mod 4. Summing 4k_i - 1 over the 25 leaves gives 4(k_1 + dots + k_25) - 25 equiv -25 equiv 3 pmod 4. So any honest total must be equiv 3 bmod 4: even a parity-passing claim like 2025 fails (2025 equiv 1 bmod 4), while 2027 would survive both screens. The smallest achievable total is leaves 1 through 25: 4(1 + dots + 25) - 25 = 1275, and from there every value equiv 3 bmod 4 is reachable by sliding one leaf index up at a time (each slide adds 4). So the honest set is exactly S ge 1275 : S equiv 3 bmod 4, capped by the ledger's length. The trap is hunting for a combination, 25 leaves from a long ledger is an astronomically large search, when the question only asks whether one can exist. The professional reflex: before reconciling a reported number line by line, test it against cheap

The Torn Ledger

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