# Puzzle 71 | Correct Number to save life

GabbarÂ andÂ hisÂ SambhaÂ areÂ fondÂ ofÂ playingÂ numberÂ games.Â ButÂ asÂ usual,Â ifÂ SambhaÂ losesÂ theÂ game,Â heÂ dies. GabbarÂ andÂ SambhaÂ takeÂ theirÂ turnÂ toÂ callÂ outÂ aÂ numberÂ betweenÂ 1Â toÂ n.Â TheyÂ followÂ theÂ followingÂ rulesÂ whileÂ playingÂ theÂ game:

• AnyoneÂ ofÂ themÂ canÂ startÂ theÂ gameÂ byÂ callingÂ aÂ numberÂ betweenÂ 1Â toÂ 10.
• TheÂ personÂ who’sÂ chanceÂ inÂ next,Â shouldÂ mustÂ callÂ aÂ numberÂ byÂ increasingÂ theÂ lastÂ numberÂ byÂ 1Â toÂ 10, Â bothÂ inclusive.

Whosoever, from Gabbar and Sambha calls out “101” first wins. Sambha tries to win this game, as it’s the run for his life. Can you help him by providing the right strategy to win the game?

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SolutionÂ :Sambha wants to call 101 first. He can do this if Gabbar calls any integer between 91 and 100 (both included), which will happen if he calls 90. He can go ahead, if Gabbar calls any integer between 80 and 89 (both included), which will happen if he calls 79. If he goes on like this, he would find out that he should call 101, 90 (=101-11), 79 (=90-11), 68 (=79-11), ….., 2.

Reverse, start with 2, let say that Gabbar calls the least one ahead, i.e. 3, then, 13, 14, 24, 25, 35, 36, 46, 47, 57, 58, 68, 69, 79, 90, 91, 101 are called by them alternatively. If incremented by 10 each time, then 101-2 falls at odd factor of 10 and Sambha starting at the odd turn will finish it at odd turn. By the above two explanations you can say that the right strategy to save his life ; he should start the game by calling 2.