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Find the exponent of 3 in 100! |
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Answer» `100!=1xx2xx3xx..xx98xx99xx100` `=(1xx2xx4xx5xx..xx98xx100)xx(3xx6xx9xx..xx96xx99)` `=Kxx3^(33)(1xx2xx3xx.xx32xx33)` `=Kxx3^(33)(1xx2xx4xx.xx31xx32)(3xx9xx12xx.xx30xx33)` `=[K(1xx2xx4xx..xx31xx32)]xx3^(33)xx(3xx9xx12xx..xx30xx33)` `=K_(1)xx3^(33)xx3^(11)(1xx2xx3xx..xx10xx11)` `=K_(1)xx(1xx2xx4xx..xx10xx11)3^(33)xx3^(11)(3xx6xx9)` `=K_(2)xx3^(33)xx3^(11)xx3^(3)xx3` `=K_(3)xx3^(48)` Hence, exponent of 3 is 48. Alternate solution : Exponent of 3 in 100! is `[(100)/(3)]+[(100)/(3^(2))]+[(100)/(3^(2))]+[(100)/(3^(4))]=33+11+3+1=48` |
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