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Question Number 157687 by MathSh last updated on 26/Oct/21

if A ; a ; b ∈ Z^+ and 6∙40!=A∙2^a ∙3^b find (a+b)_(max) = ?

$$\mathrm{if}\:\:\:\mathrm{A}\:;\:\mathrm{a}\:;\:\mathrm{b}\:\in\:\mathbb{Z}^{+} \:\:\mathrm{and}\:\:\mathrm{6}\centerdot\mathrm{40}!=\mathrm{A}\centerdot\mathrm{2}^{\boldsymbol{\mathrm{a}}} \centerdot\mathrm{3}^{\boldsymbol{\mathrm{b}}} \\ $$$$\mathrm{find}\:\:\:\left(\mathrm{a}+\mathrm{b}\right)_{\boldsymbol{\mathrm{max}}} \:=\:? \\ $$

Commented by Rasheed.Sindhi last updated on 26/Oct/21

(a+b)_(max) = 58

$$\left(\mathrm{a}+\mathrm{b}\right)_{\boldsymbol{\mathrm{max}}} \:=\:\mathrm{58} \\ $$

Answered by Rasheed.Sindhi last updated on 26/Oct/21

Account of factor 2 in 6.40!: •2,4,6,...,40(counting one factor) 2^(20) •4,8,12,...,40(counting 2nd factor) 2^(10) •8,16,...,40(counting 3rd factor) 2^5 •16,32 (counting 4th factor) 2^2 • 32(counting 5th factor) 2^1 • 6: 2^1 Maximum factors of 2: 2^(39) Account of factor 3 in 6.40!: • 3,6,9,...39(13 numbers): 3^(13) (counting only one factor) • 9,18,27,36 :3^4 (counting second factor of each) • 27(counting 3rd factor) 3^1 • 6: 3^1 Maximum number of factors:3^(19) (a+b)_(max) =39+19=58

$${Account}\:{of}\:{factor}\:\mathrm{2}\:{in}\:\mathrm{6}.\mathrm{40}!: \\ $$$$\bullet\mathrm{2},\mathrm{4},\mathrm{6},...,\mathrm{40}\left({counting}\:{one}\:{factor}\right)\:\mathrm{2}^{\mathrm{20}} \\ $$$$\bullet\mathrm{4},\mathrm{8},\mathrm{12},...,\mathrm{40}\left({counting}\:\mathrm{2}{nd}\:{factor}\right)\:\mathrm{2}^{\mathrm{10}} \\ $$$$\bullet\mathrm{8},\mathrm{16},...,\mathrm{40}\left({counting}\:\mathrm{3}{rd}\:{factor}\right)\:\mathrm{2}^{\mathrm{5}} \\ $$$$\bullet\mathrm{16},\mathrm{32}\:\left({counting}\:\mathrm{4}{th}\:{factor}\right)\:\:\:\mathrm{2}^{\mathrm{2}} \\ $$$$\bullet\:\mathrm{32}\left({counting}\:\mathrm{5}{th}\:{factor}\right)\:\mathrm{2}^{\mathrm{1}} \\ $$$$\bullet\:\mathrm{6}:\:\mathrm{2}^{\mathrm{1}} \\ $$$${Maximum}\:{factors}\:{of}\:\mathrm{2}:\:\mathrm{2}^{\mathrm{39}} \\ $$$${Account}\:{of}\:{factor}\:\mathrm{3}\:{in}\:\mathrm{6}.\mathrm{40}!: \\ $$$$\bullet\:\:\mathrm{3},\mathrm{6},\mathrm{9},...\mathrm{39}\left(\mathrm{13}\:{numbers}\right):\:\mathrm{3}^{\mathrm{13}} \\ $$$$\:\:\:\:\left({counting}\:{only}\:{one}\:{factor}\right) \\ $$$$\bullet\:\mathrm{9},\mathrm{18},\mathrm{27},\mathrm{36}\:\:\:\::\mathrm{3}^{\mathrm{4}} \\ $$$$\left({counting}\:{second}\:{factor}\:{of}\:{each}\right) \\ $$$$\bullet\:\mathrm{27}\left({counting}\:\mathrm{3}{rd}\:{factor}\right)\:\:\mathrm{3}^{\mathrm{1}} \\ $$$$\bullet\:\mathrm{6}:\:\:\:\:\mathrm{3}^{\mathrm{1}} \\ $$$${Maximum}\:{number}\:{of}\:{factors}:\mathrm{3}^{\mathrm{19}} \\ $$$$\left({a}+{b}\right)_{{max}} =\mathrm{39}+\mathrm{19}=\mathrm{58} \\ $$

Commented by MathSh last updated on 26/Oct/21

Very nice dear Ser, thank you so much

$$\mathrm{Very}\:\mathrm{nice}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

Answered by mr W last updated on 26/Oct/21

⌊((40)/2)⌋+⌊((40)/2^2 )⌋+⌊((40)/2^3 )⌋+⌊((40)/2^4 )⌋+⌊((40)/2^5 )⌋=20+10+5+2+1=38 ⇒40! contains 2^(38) ⌊((40)/3)⌋+⌊((40)/3^2 )⌋+⌊((40)/3^3 )⌋=13+4+1=18 ⇒40! contains 3^(18) 6=2^1 ×3^1 ⇒6×40!=2^(39) ×3^(19) ×n with gcd(n,2)=1, gcd(n,3)=1 therefore for 6×40!=A×2^a ×3^b , a_(max) =39, b_(max) =19 (a+b)_(max) =39+19=58

$$\lfloor\frac{\mathrm{40}}{\mathrm{2}}\rfloor+\lfloor\frac{\mathrm{40}}{\mathrm{2}^{\mathrm{2}} }\rfloor+\lfloor\frac{\mathrm{40}}{\mathrm{2}^{\mathrm{3}} }\rfloor+\lfloor\frac{\mathrm{40}}{\mathrm{2}^{\mathrm{4}} }\rfloor+\lfloor\frac{\mathrm{40}}{\mathrm{2}^{\mathrm{5}} }\rfloor=\mathrm{20}+\mathrm{10}+\mathrm{5}+\mathrm{2}+\mathrm{1}=\mathrm{38} \\ $$$$\Rightarrow\mathrm{40}!\:{contains}\:\mathrm{2}^{\mathrm{38}} \\ $$$$\lfloor\frac{\mathrm{40}}{\mathrm{3}}\rfloor+\lfloor\frac{\mathrm{40}}{\mathrm{3}^{\mathrm{2}} }\rfloor+\lfloor\frac{\mathrm{40}}{\mathrm{3}^{\mathrm{3}} }\rfloor=\mathrm{13}+\mathrm{4}+\mathrm{1}=\mathrm{18} \\ $$$$\Rightarrow\mathrm{40}!\:{contains}\:\mathrm{3}^{\mathrm{18}} \\ $$$$\mathrm{6}=\mathrm{2}^{\mathrm{1}} ×\mathrm{3}^{\mathrm{1}} \\ $$$$\Rightarrow\mathrm{6}×\mathrm{40}!=\mathrm{2}^{\mathrm{39}} ×\mathrm{3}^{\mathrm{19}} ×{n}\: \\ $$$${with}\:{gcd}\left({n},\mathrm{2}\right)=\mathrm{1},\:{gcd}\left({n},\mathrm{3}\right)=\mathrm{1} \\ $$$${therefore}\:{for}\:\mathrm{6}×\mathrm{40}!={A}×\mathrm{2}^{{a}} ×\mathrm{3}^{{b}} , \\ $$$${a}_{{max}} =\mathrm{39},\:{b}_{{max}} =\mathrm{19} \\ $$$$\left({a}+{b}\right)_{{max}} =\mathrm{39}+\mathrm{19}=\mathrm{58} \\ $$

Commented by MathSh last updated on 26/Oct/21

perfect dear Ser thank you

$$\mathrm{perfect}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thank}\:\mathrm{you} \\ $$

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