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Question Number 5824 by Bc160200609 last updated on 30/May/16

The exponent of  12 in  100!  is

$$\mathrm{The}\:\mathrm{exponent}\:\mathrm{of}\:\:\mathrm{12}\:\mathrm{in}\:\:\mathrm{100}!\:\:\mathrm{is}\: \\ $$

Commented by Yozzii last updated on 30/May/16

12^(48) =2^(96) 3^(48)  but I haven′t figured  out yet how to show this.  The isolation of powers of 2 and 3  in the product of 100! is required.  12=2^2 ×3. There may be a simple  method...

$$\mathrm{12}^{\mathrm{48}} =\mathrm{2}^{\mathrm{96}} \mathrm{3}^{\mathrm{48}} \:{but}\:{I}\:{haven}'{t}\:{figured} \\ $$$${out}\:{yet}\:{how}\:{to}\:{show}\:{this}. \\ $$$${The}\:{isolation}\:{of}\:{powers}\:{of}\:\mathrm{2}\:{and}\:\mathrm{3} \\ $$$${in}\:{the}\:{product}\:{of}\:\mathrm{100}!\:{is}\:{required}. \\ $$$$\mathrm{12}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}.\:{There}\:{may}\:{be}\:{a}\:{simple} \\ $$$${method}... \\ $$

Commented by prakash jain last updated on 30/May/16

Exponent of 2=⌊((100)/2)⌋+⌊((100)/2^2 )⌋+...+⌊((100)/2^6 )⌋  =50+25+12+6+3+1=97  Exponent of 3=⌊((100)/3)⌋+⌊((100)/3^2 )⌋+⌊((100)/3^3 )⌋+⌊((100)/3^4 )⌋  =33+11+3+1=48  12=2^2 ×3  So highest power of 12=48  12^(48) =2^(96) ×3^(48)

$$\mathrm{Exponent}\:\mathrm{of}\:\mathrm{2}=\lfloor\frac{\mathrm{100}}{\mathrm{2}}\rfloor+\lfloor\frac{\mathrm{100}}{\mathrm{2}^{\mathrm{2}} }\rfloor+...+\lfloor\frac{\mathrm{100}}{\mathrm{2}^{\mathrm{6}} }\rfloor \\ $$$$=\mathrm{50}+\mathrm{25}+\mathrm{12}+\mathrm{6}+\mathrm{3}+\mathrm{1}=\mathrm{97} \\ $$$$\mathrm{Exponent}\:\mathrm{of}\:\mathrm{3}=\lfloor\frac{\mathrm{100}}{\mathrm{3}}\rfloor+\lfloor\frac{\mathrm{100}}{\mathrm{3}^{\mathrm{2}} }\rfloor+\lfloor\frac{\mathrm{100}}{\mathrm{3}^{\mathrm{3}} }\rfloor+\lfloor\frac{\mathrm{100}}{\mathrm{3}^{\mathrm{4}} }\rfloor \\ $$$$=\mathrm{33}+\mathrm{11}+\mathrm{3}+\mathrm{1}=\mathrm{48} \\ $$$$\mathrm{12}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3} \\ $$$$\mathrm{So}\:\mathrm{highest}\:\mathrm{power}\:\mathrm{of}\:\mathrm{12}=\mathrm{48} \\ $$$$\mathrm{12}^{\mathrm{48}} =\mathrm{2}^{\mathrm{96}} ×\mathrm{3}^{\mathrm{48}} \\ $$

Commented by Yozzii last updated on 30/May/16

Is there an explanation for the  forumula?

$${Is}\:{there}\:{an}\:{explanation}\:{for}\:{the} \\ $$$${forumula}? \\ $$

Commented by prakash jain last updated on 01/Jun/16

Every multiple of 2 in 1 to 100 adds 1 to  exponent of 100!  Every multiple of 4 in 1−100 adds 2 to  exponent of 100!  and so on.  so exponent of 2 in 100!  =⌊((100)/2)⌋+⌊((100)/4)⌋+⌊((100)/8)⌋+⌊((100)/(16))⌋+⌊((100)/(32))⌋+⌊((100)/(64))⌋  =50+25+12+6+3+1=97

$$\mathrm{Every}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{2}\:\mathrm{in}\:\mathrm{1}\:\mathrm{to}\:\mathrm{100}\:\mathrm{adds}\:\mathrm{1}\:\mathrm{to} \\ $$$$\mathrm{exponent}\:\mathrm{of}\:\mathrm{100}! \\ $$$$\mathrm{Every}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{4}\:\mathrm{in}\:\mathrm{1}−\mathrm{100}\:\mathrm{adds}\:\mathrm{2}\:\mathrm{to} \\ $$$$\mathrm{exponent}\:\mathrm{of}\:\mathrm{100}! \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{on}. \\ $$$$\mathrm{so}\:\mathrm{exponent}\:\mathrm{of}\:\mathrm{2}\:\mathrm{in}\:\mathrm{100}! \\ $$$$=\lfloor\frac{\mathrm{100}}{\mathrm{2}}\rfloor+\lfloor\frac{\mathrm{100}}{\mathrm{4}}\rfloor+\lfloor\frac{\mathrm{100}}{\mathrm{8}}\rfloor+\lfloor\frac{\mathrm{100}}{\mathrm{16}}\rfloor+\lfloor\frac{\mathrm{100}}{\mathrm{32}}\rfloor+\lfloor\frac{\mathrm{100}}{\mathrm{64}}\rfloor \\ $$$$=\mathrm{50}+\mathrm{25}+\mathrm{12}+\mathrm{6}+\mathrm{3}+\mathrm{1}=\mathrm{97} \\ $$

Commented by Yozzii last updated on 01/Jun/16

I see. Thanks!

$${I}\:{see}.\:{Thanks}!\: \\ $$

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