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Question Number 156775 by mnjuly1970 last updated on 15/Oct/21

f is a function that assigns to each natural number multiplication of its digits for exampl : f( 200)=0 f ( 128 )=1×2×8 =16 find the value of : f (100) + f(101 )+ f (102 )+...f(999)=?

$$ \\ $$$$\:\:\:{f}\:\:{is}\:\:{a}\:{function}\:{that}\:{assigns} \\ $$$$\:\:\:{to}\:{each}\:{natural}\:{number}\: \\ $$$$\:\:\:\:\:{multiplication}\:{of}\:{its}\:{digits}\: \\ $$$$\:\:\:\:{for}\:{exampl}\::\:\:{f}\left(\:\mathrm{200}\right)=\mathrm{0} \\ $$$$\:\:\:\:{f}\:\left(\:\mathrm{128}\:\right)=\mathrm{1}×\mathrm{2}×\mathrm{8}\:=\mathrm{16} \\ $$$$\:\:\:\:\:{find}\:{the}\:{value}\:{of}\:: \\ $$$$\:\:\:\:\:{f}\:\left(\mathrm{100}\right)\:+\:{f}\left(\mathrm{101}\:\right)+\:{f}\:\left(\mathrm{102}\:\right)+...{f}\left(\mathrm{999}\right)=? \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 15/Oct/21

Σ_(a=1,b=0,c=0) ^(a=9,b=9,c=9) abc=91125

$$\underset{{a}=\mathrm{1},{b}=\mathrm{0},{c}=\mathrm{0}} {\overset{{a}=\mathrm{9},{b}=\mathrm{9},{c}=\mathrm{9}} {\sum}}{abc}=\mathrm{91125} \\ $$

Answered by Rasheed.Sindhi last updated on 15/Oct/21

100≤abc^(−) ≤999 c=0,1,2,...,9 ab0^(−) ,ab1^(−) ,ab2^(−) ,...ab9^(−) f(ab0^(−) )+f(ab1^(−) )+f(ab2^(−) )+,...+f(ab9^(−) ) =a.b.0+a.b.1+a.b.2+...+a.b.9 =ab(0+1+2+3+...+9) =ab(((9×10)/2))=ab(45) b=0,1,2,...,9 a.b(45): 45a.0+45a.1+45a.2+...+45a.9 =45a(0+1+2+...+9)=45^2 a a=1,2,3,...,9 45^2 a: 45^2 .1+45^2 .2+45^2 .3+...+45^2 .9 45^2 (1+2+3+...+9)=45^3 =91125 By my son, Arshad Soomro

$$\mathrm{100}\leqslant\overline {{abc}}\leqslant\mathrm{999} \\ $$$${c}=\mathrm{0},\mathrm{1},\mathrm{2},...,\mathrm{9} \\ $$$$\overline {{ab}\mathrm{0}},\overline {{ab}\mathrm{1}},\overline {{ab}\mathrm{2}},...\overline {{ab}\mathrm{9}} \\ $$$${f}\left(\overline {{ab}\mathrm{0}}\right)+{f}\left(\overline {{ab}\mathrm{1}}\right)+{f}\left(\overline {{ab}\mathrm{2}}\right)+,...+{f}\left(\overline {{ab}\mathrm{9}}\right) \\ $$$$={a}.{b}.\mathrm{0}+{a}.{b}.\mathrm{1}+{a}.{b}.\mathrm{2}+...+{a}.{b}.\mathrm{9} \\ $$$$={ab}\left(\mathrm{0}+\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{9}\right) \\ $$$$={ab}\left(\frac{\mathrm{9}×\mathrm{10}}{\mathrm{2}}\right)={ab}\left(\mathrm{45}\right) \\ $$$${b}=\mathrm{0},\mathrm{1},\mathrm{2},...,\mathrm{9} \\ $$$${a}.{b}\left(\mathrm{45}\right): \\ $$$$\:\mathrm{45}{a}.\mathrm{0}+\mathrm{45}{a}.\mathrm{1}+\mathrm{45}{a}.\mathrm{2}+...+\mathrm{45}{a}.\mathrm{9} \\ $$$$=\mathrm{45}{a}\left(\mathrm{0}+\mathrm{1}+\mathrm{2}+...+\mathrm{9}\right)=\mathrm{45}^{\mathrm{2}} {a} \\ $$$${a}=\mathrm{1},\mathrm{2},\mathrm{3},...,\mathrm{9} \\ $$$$\mathrm{45}^{\mathrm{2}} {a}: \\ $$$$\mathrm{45}^{\mathrm{2}} .\mathrm{1}+\mathrm{45}^{\mathrm{2}} .\mathrm{2}+\mathrm{45}^{\mathrm{2}} .\mathrm{3}+...+\mathrm{45}^{\mathrm{2}} .\mathrm{9} \\ $$$$\mathrm{45}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{2}+\mathrm{3}+...+\mathrm{9}\right)=\mathrm{45}^{\mathrm{3}} =\mathrm{91125} \\ $$$$\:\:\:\:\:\:\:\mathcal{B}{y}\:{my}\:{son},\:\mathrm{Arshad}\:\mathrm{Soomro} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 16/Oct/21

very nice thanks alot ...

$${very}\:{nice}\:{thanks}\:{alot}\:... \\ $$

Commented by qaz last updated on 15/Oct/21

S=Σ_(a=1) ^9 Σ_(b=1) ^9 Σ_(c=1) ^9 abc=Σ_(a=1) ^9 aΣ_(b=1) ^9 bΣ_(c=1) ^9 c=(((9∙10)/2))^3 =91125

$$\mathrm{S}=\underset{\mathrm{a}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\underset{\mathrm{b}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\underset{\mathrm{c}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\mathrm{abc}=\underset{\mathrm{a}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\mathrm{a}\underset{\mathrm{b}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\mathrm{b}\underset{\mathrm{c}=\mathrm{1}} {\overset{\mathrm{9}} {\sum}}\mathrm{c}=\left(\frac{\mathrm{9}\centerdot\mathrm{10}}{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{91125} \\ $$

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