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5-555-5-50-find-the-sum-of-the-digits-of-the-product-




Question Number 201557 by hardmath last updated on 08/Dec/23
5 ∙ 555...5_( 50)   find the sum of the digits of the  product.
$$\mathrm{5}\:\centerdot\:\underset{\:\mathrm{50}} {\underbrace{\mathrm{555}…\mathrm{5}}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{product}. \\ $$
Answered by aleks041103 last updated on 09/Dec/23
5.5=25  5.55=5.50+5.5=250+25=275  5.555=5.500+5.55=2500+275=2775  ...  5.555...5=277...775  in this case  5 ∙ 555...5_( 50)  = 27...7_(49)  5  ⇒sum of digits is 7.50=350
$$\mathrm{5}.\mathrm{5}=\mathrm{25} \\ $$$$\mathrm{5}.\mathrm{55}=\mathrm{5}.\mathrm{50}+\mathrm{5}.\mathrm{5}=\mathrm{250}+\mathrm{25}=\mathrm{275} \\ $$$$\mathrm{5}.\mathrm{555}=\mathrm{5}.\mathrm{500}+\mathrm{5}.\mathrm{55}=\mathrm{2500}+\mathrm{275}=\mathrm{2775} \\ $$$$… \\ $$$$\mathrm{5}.\mathrm{555}…\mathrm{5}=\mathrm{277}…\mathrm{775} \\ $$$${in}\:{this}\:{case} \\ $$$$\mathrm{5}\:\centerdot\:\underset{\:\mathrm{50}} {\underbrace{\mathrm{555}…\mathrm{5}}}\:=\:\mathrm{2}\underset{\mathrm{49}} {\underbrace{\mathrm{7}…\mathrm{7}}}\:\mathrm{5} \\ $$$$\Rightarrow{sum}\:{of}\:{digits}\:{is}\:\mathrm{7}.\mathrm{50}=\mathrm{350} \\ $$
Commented by hardmath last updated on 09/Dec/23
cool dear professor thankyou
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thankyou} \\ $$
Answered by mr W last updated on 09/Dec/23
5×555...5_(n times)   =25×111...1_(n times)   =25         25            25               ...                  25  =2777...7_((n−1) times) 5  sum of digits =2+7(n−1)+5=7n  with n=50 we have 7×50=350 ✓
$$\mathrm{5}×\underset{{n}\:{times}} {\mathrm{555}…\mathrm{5}} \\ $$$$=\mathrm{25}×\underset{{n}\:{times}} {\mathrm{111}…\mathrm{1}} \\ $$$$=\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:… \\ $$$$\underline{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{25}} \\ $$$$=\mathrm{2}\underset{\left({n}−\mathrm{1}\right)\:{times}} {\mathrm{777}…\mathrm{7}5} \\ $$$${sum}\:{of}\:{digits}\:=\mathrm{2}+\mathrm{7}\left({n}−\mathrm{1}\right)+\mathrm{5}=\mathrm{7}{n} \\ $$$${with}\:{n}=\mathrm{50}\:{we}\:{have}\:\mathrm{7}×\mathrm{50}=\mathrm{350}\:\checkmark \\ $$
Commented by hardmath last updated on 09/Dec/23
cool dear professor thankyou
$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thankyou} \\ $$

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