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Category: Number Theory

Prove-that-if-two-numbers-are-chosen-at-random-then-the-probability-that-their-sum-is-divisible-by-n-is-1-n-

Question Number 785 by rishabh last updated on 12/Mar/15 $${Prove}\:{that}\:{if}\:{two}\:{numbers}\:{are}\:{chosen} \\ $$$${at}\:{random}\:{then}\:{the}\:{probability}\:{that} \\ $$$${their}\:{sum}\:{is}\:{divisible}\:{by}\:{n}\:{is}\:\frac{\mathrm{1}}{{n}}. \\ $$ Answered by prakash jain last updated on 12/Mar/15 $$\mathrm{Sum}\:\mathrm{mod}\:{n}={k},\:\mathrm{where}\:\mathrm{0}\leqslant{k}\leqslant{n}−\mathrm{1}.…

List-all-primes-p-for-which-p-2-and-p-4-are-also-primes-

Question Number 765 by prakash jain last updated on 09/Mar/15 $$\mathrm{List}\:\mathrm{all}\:\mathrm{primes}\:{p}\:\mathrm{for}\:\mathrm{which}\:{p}+\mathrm{2}\:\mathrm{and}\:{p}+\mathrm{4} \\ $$$$\mathrm{are}\:\mathrm{also}\:\mathrm{primes}. \\ $$ Answered by rishabh last updated on 09/Mar/15 $${p}=\mathrm{3}\:\mathrm{is}\:\mathrm{the}\:\mathrm{only}\:\mathrm{such}\:\mathrm{number}. \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{assume}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:{p}>\mathrm{3}.…

Prove-that-for-p-q-r-N-lcm-p-q-r-gcd-p-q-r-p-q-r-gcd-p-q-gcd-q-r-gcd-r-p-

Question Number 66275 by Rasheed.Sindhi last updated on 12/Aug/19 $${Prove}\:{that}\:{for}\:{p},{q},{r}\in\mathbb{N} \\ $$$$\frac{\mathrm{lcm}\left({p},{q},{r}\right)}{\mathrm{gcd}\left({p},{q},{r}\right)}=\frac{{p}×{q}×{r}}{\mathrm{gcd}\left({p},{q}\right)×\mathrm{gcd}\left({q},{r}\right)×\mathrm{gcd}\left({r},{p}\right)} \\ $$ Commented by Prithwish sen last updated on 12/Aug/19 $$\mathrm{p}=\mathrm{nn}_{\mathrm{1}} \mathrm{n}_{\mathrm{2}} \mathrm{x},\mathrm{q}=\mathrm{nn}_{\mathrm{2}}…

Prove-by-induction-on-n-for-n-2-u-n-2-3-n-1-for-the-sequence-u-n-defined-by-the-recurrence-relation-u-1-1-

Question Number 591 by 112358 last updated on 04/Feb/15 $${Prove}\:{by}\:{induction}\:{on}\:{n},\:{for}\:{n}\geqslant\mathrm{2}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{u}_{{n}} \:\geqslant\:\mathrm{2}^{\mathrm{3}^{{n}−\mathrm{1}} } \\ $$$${for}\:{the}\:{sequence}\:\left\{{u}_{{n}} \right\}\:{defined}\:{by}\: \\ $$$${the}\:{recurrence}\:{relation} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{u}_{\mathrm{1}} =\mathrm{1}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{u}_{{n}+\mathrm{1}} =\left({u}_{{n}}…