Race to Sixty

Question

Two clerks play a counting duel. Starting from zero, they alternately add any integer from 1 to 7 to a running total and announce the new total; the total may never pass 60, and whoever announces exactly 60 wins the pot. You move first. What total do you announce, and how do you hold the win from there?

Accepted Answer

Announce 4. Afterwards, whenever your opponent adds k, you add 8 - k. The ladder. Work backward from the target. From totals 53 through 59 the mover wins on the spot, so 52 is the total you want to hand over: every move from 52 lands in 53..59, and you finish. The same logic repeats every 8: the losing totals (for the player who must move) are 4, 12, 20, 28, 36, 44, 52, 60 quad text(totals equiv 4 bmod 8text). Your opening announcement of 4 puts your opponent on the bottom rung. Because a single move adds between 1 and 7, never 8, your opponent can't hop from one rung to the next, while your 8-k reply always restores the invariant "total equiv 4 bmod 8 when they move." The ladder climbs 4 to 12 to dots to 52, and from 52 their move hands you 60. An exhaustive backward solve of all 60 totals confirms 4 is the only winning first announcement. Why 8 and why 4: the modulus is 1 + 7, the smallest and largest step glued together, the pair sum you can always complete; the residue is 60 bmod 8 = 4. That two-line recipe (m = max + min steps... for step sets 1..m the modulus is m+1, residue = N bmod (m+1)) solves the entire race-to-N family instantly. The trap: anchoring only on "say 52" and

Race to Sixty

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