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Question Number 208071 by hardmath last updated on 03/Jun/24

$$\frac{4\cdot {10}^{\mathbf{n}}-52}{3}$$$$\mathrm{The}\mathrm{numbers}\mathrm{are}38\mathrm{in}\mathrm{total}$$$$\mathrm{Find}:\mathbf{n}=?$$

Commented by mr W last updated on 04/Jun/24

$${it}^{\prime }snotclearwhatyoumean.$$

Commented by hardmath last updated on 04/Jun/24

$$\mathrm{dear}\mathrm{professor},\mathrm{answer}:12$$

Commented by mr W last updated on 04/Jun/24

$$howisananswerpossible,ifitis$$$$notclearwhatthequestionis?$$

Commented by TonyCWX08 last updated on 04/Jun/24

$$Sinceyouhavetheanswer,whyareyouhere?$$

Commented by Frix last updated on 04/Jun/24

$$\mathrm{Maybe}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{digits}\mathrm{is}38?$$$$\mathrm{In}\mathrm{this}\mathrm{case}:$$$$n\mathrm{Sum}\mathrm{of}\mathrm{digits}\mathrm{of}\frac{4\times {10}^n-52}{3}$$$$21+1+6$$$$31+3+1+6$$$$41+3+3+1+6$$$$n1+(n-2)\times 3+1+6=3n+2$$$$3n+2=38$$$$n=12$$$$\frac{4\times {10}^{12}-52}{3}=\underset{\mathrm{Sum}=38}{\underbrace{1333333333316}}$$

Commented by hardmath last updated on 04/Jun/24

$$\mathrm{yes}\mathrm{professor}$$

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