Category: Arithmetic

Question Number 193280 by Mingma last updated on 09/Jun/23 Answered by AST last updated on 09/Jun/23 $$1000\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}{def}+\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}{abc}=6000\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}{abc}+6\stackrel{\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}}{def}\Rightarrow \frac{def}{abc}=\frac{5999}{994}=\frac{857}{142}$$$$\Rightarrow\overset{\_\_\_\_\_\_\_\_} {{abcdef}}=\mathrm{142857} \…

Question Number 131054 by benjo_mathlover last updated on 01/Feb/21 $$\mathrm{Find}\mathrm{the}\mathrm{first}\mathrm{four}\mathrm{terms}\mathrm{in}\mathrm{the}\mathrm{series}$$$$\mathrm{expansion}\mathrm{of}\frac{1}{3\mathrm{x}+5}\mathrm{ascending}\mathrm{power}$$$$\mathrm{x}\mathrm{and}\mathrm{state}\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{for}$$$$\mathrm{which}\mathrm{this}\mathrm{expansion}\mathrm{is}\mathrm{valid}.$$ Answered by EDWIN88 last updated on 01/Feb/21…

Question Number 130936 by mathlove last updated on 30/Jan/21 Answered by floor(10²Eta[1]) last updated on 30/Jan/21 $$15\equiv 1\left(\mathrm{mod}7\right)$$$$\mathrm{so}{15}^{\mathrm{n}}\equiv 1^{\mathrm{n}}=1\left(\mathrm{mod}7\right)$$$$\Rightarrow\mathrm{3}^{\mathrm{3}^{\mathrm{2007}} } ×\mathrm{15}^{\mathrm{2}^{\mathrm{7777}}…

Question Number 65199 by rajesh4661kumar@gmail.com last updated on 26/Jul/19 Answered by Tanmay chaudhury last updated on 26/Jul/19 $$\left(1\right),(2,3,4),(5,6,7,8,9),(10,11,12,13,14,15,16)\dots$$$$1stgroupcontains\to 1terms$$$$2ndgroupcontains\to 3terms$$$$\mathrm{3}{rd}\:\:\:{group}\:\:\:{contains}\rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}{terms} \…

Question Number 64955 by Tony Lin last updated on 23/Jul/19 $$provethat\underset{k=1}{\sum ^n}k\cdot \frac{(k+1)!}{2^{k+1}}=\frac{(n+2)!}{2^{n+1}}-1$$ Commented by ~ À ® @ 237 ~…