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Question Number 65214 by naka3546 last updated on 26/Jul/19

Prove or disprove that 2^(101) ∣ n^n − 101 .

$${Prove}\:\:{or}\:\:{disprove}\:\:\:{that}\:\:\:\mathrm{2}^{\mathrm{101}} \:\mid\:{n}^{{n}} \:−\:\mathrm{101}\:. \\ $$

Commented by naka3546 last updated on 26/Jul/19

n positive integer

$${n}\:\:{positive}\:\:{integer} \\ $$

Commented by Rasheed.Sindhi last updated on 27/Jul/19

For n=1, n^n −101=1^1 −101=−100 And 2^(101) ∤ −100.

$${For}\:{n}=\mathrm{1},\:{n}^{{n}} −\mathrm{101}=\mathrm{1}^{\mathrm{1}} −\mathrm{101}=−\mathrm{100} \\ $$$${And}\:\mathrm{2}^{\mathrm{101}} \nmid\:−\mathrm{100}. \\ $$

Commented by naka3546 last updated on 27/Jul/19

any n that satisfy on ?

$${any}\:\:{n}\:\:{that}\:\:{satisfy}\:\:{on}\:? \\ $$

Answered by MJS last updated on 27/Jul/19

n^n −101≥2^(101) ⇒ n≥2^(101) +101=2 535 301 200 456 458 802 993 406 410 853 n^n −101=m×2^(101) n^n −101=2m×2^(100) n^n −101=2l n^n =2l+101 n^n =2(l+50)+1 n^n =2k+1 ⇒ it′s not true for even n 2^(101) −101=7^2 ×263 201×196 582 994 842 884 397 808 597= =p^2 qr n^n =m×p^2 qr m=(n^n /(p^2 qr))=(((p^α q^β r^γ )^((p^α q^β r^γ )) )/(p^2 qr))= =p^(αp^α q^β r^γ −2) q^(βp^α q^β r^γ −1) r^(γp^α q^β r^γ −1) the smallest number m is the one with α=β=γ=1 m_(min) =p^(pqr−2) q^(pqr−1) r^(pqr−1) ≈10^((10^(31) ))

$${n}^{{n}} −\mathrm{101}\geqslant\mathrm{2}^{\mathrm{101}} \:\Rightarrow\:{n}\geqslant\mathrm{2}^{\mathrm{101}} +\mathrm{101}=\mathrm{2}\:\mathrm{535}\:\mathrm{301}\:\mathrm{200}\:\mathrm{456}\:\mathrm{458}\:\mathrm{802}\:\mathrm{993}\:\mathrm{406}\:\mathrm{410}\:\mathrm{853} \\ $$$$ \\ $$$${n}^{{n}} −\mathrm{101}={m}×\mathrm{2}^{\mathrm{101}} \\ $$$${n}^{{n}} −\mathrm{101}=\mathrm{2}{m}×\mathrm{2}^{\mathrm{100}} \\ $$$${n}^{{n}} −\mathrm{101}=\mathrm{2}{l} \\ $$$${n}^{{n}} =\mathrm{2}{l}+\mathrm{101} \\ $$$${n}^{{n}} =\mathrm{2}\left({l}+\mathrm{50}\right)+\mathrm{1} \\ $$$${n}^{{n}} =\mathrm{2}{k}+\mathrm{1}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{true}\:\mathrm{for}\:\mathrm{even}\:{n} \\ $$$$ \\ $$$$\mathrm{2}^{\mathrm{101}} −\mathrm{101}=\mathrm{7}^{\mathrm{2}} ×\mathrm{263}\:\mathrm{201}×\mathrm{196}\:\mathrm{582}\:\mathrm{994}\:\mathrm{842}\:\mathrm{884}\:\mathrm{397}\:\mathrm{808}\:\mathrm{597}= \\ $$$$={p}^{\mathrm{2}} {qr} \\ $$$${n}^{{n}} ={m}×{p}^{\mathrm{2}} {qr} \\ $$$${m}=\frac{{n}^{{n}} }{{p}^{\mathrm{2}} {qr}}=\frac{\left({p}^{\alpha} {q}^{\beta} {r}^{\gamma} \right)^{\left({p}^{\alpha} {q}^{\beta} {r}^{\gamma} \right)} }{{p}^{\mathrm{2}} {qr}}= \\ $$$$={p}^{\alpha{p}^{\alpha} {q}^{\beta} {r}^{\gamma} −\mathrm{2}} {q}^{\beta{p}^{\alpha} {q}^{\beta} {r}^{\gamma} −\mathrm{1}} {r}^{\gamma{p}^{\alpha} {q}^{\beta} {r}^{\gamma} −\mathrm{1}} \\ $$$$\mathrm{the}\:\mathrm{smallest}\:\mathrm{number}\:{m}\:\mathrm{is}\:\mathrm{the}\:\mathrm{one}\:\mathrm{with} \\ $$$$\alpha=\beta=\gamma=\mathrm{1} \\ $$$${m}_{{min}} ={p}^{{pqr}−\mathrm{2}} {q}^{{pqr}−\mathrm{1}} {r}^{{pqr}−\mathrm{1}} \approx\mathrm{10}^{\left(\mathrm{10}^{\mathrm{31}} \right)} \\ $$

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