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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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