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Question Number 162118 by SANOGO last updated on 27/Dec/21

determiner le reste de la division euclidienne de 10^(100) par 105

$${determiner}\:{le}\:{reste}\:{de}\:{la}\:{division}\:{euclidienne}\:{de} \\ $$$$\mathrm{10}^{\mathrm{100}} \:{par}\:\mathrm{105} \\ $$

Commented by SANOGO last updated on 27/Dec/21

thank you

$${thank}\:{you} \\ $$

Commented by SANOGO last updated on 27/Dec/21

merci bien

$${merci}\:{bien} \\ $$

Commented by mr W last updated on 27/Dec/21

in following stupid method with “=^R ” i mean “has the same remainder as”. 10^(100) =10×10^(99) =10×(1000)^(33) =10×(9×105+55)^(33) =^R 10×55^(33) =10×55×(3025)^(16) =10×55×(28×105+85)^(16) =^R 10×55×85^(16) =10×55×(68×105+85)^8 =^R 10×55×85^8 =^R 10×55×85^4 =^R 10×55×85^2 =^R 10×55×85 =^R 25 i.e. the remainder is 25 when 10^(100) is divided by 105.

$${in}\:{following}\:{stupid}\:{method}\: \\ $$$${with}\:``\overset{{R}} {=}''\:{i}\:{mean} \\ $$$$``{has}\:{the}\:{same}\:{remainder}\:{as}''. \\ $$$$\mathrm{10}^{\mathrm{100}} =\mathrm{10}×\mathrm{10}^{\mathrm{99}} =\mathrm{10}×\left(\mathrm{1000}\right)^{\mathrm{33}} \\ $$$$=\mathrm{10}×\left(\mathrm{9}×\mathrm{105}+\mathrm{55}\right)^{\mathrm{33}} \\ $$$$\overset{{R}} {=}\mathrm{10}×\mathrm{55}^{\mathrm{33}} =\mathrm{10}×\mathrm{55}×\left(\mathrm{3025}\right)^{\mathrm{16}} \\ $$$$=\mathrm{10}×\mathrm{55}×\left(\mathrm{28}×\mathrm{105}+\mathrm{85}\right)^{\mathrm{16}} \\ $$$$\overset{{R}} {=}\mathrm{10}×\mathrm{55}×\mathrm{85}^{\mathrm{16}} =\mathrm{10}×\mathrm{55}×\left(\mathrm{68}×\mathrm{105}+\mathrm{85}\right)^{\mathrm{8}} \\ $$$$\overset{{R}} {=}\mathrm{10}×\mathrm{55}×\mathrm{85}^{\mathrm{8}} \\ $$$$\overset{{R}} {=}\mathrm{10}×\mathrm{55}×\mathrm{85}^{\mathrm{4}} \\ $$$$\overset{{R}} {=}\mathrm{10}×\mathrm{55}×\mathrm{85}^{\mathrm{2}} \\ $$$$\overset{{R}} {=}\mathrm{10}×\mathrm{55}×\mathrm{85} \\ $$$$\overset{{R}} {=}\mathrm{25} \\ $$$${i}.{e}.\:{the}\:{remainder}\:{is}\:\mathrm{25}\:{when}\:\mathrm{10}^{\mathrm{100}} \\ $$$${is}\:{divided}\:{by}\:\mathrm{105}. \\ $$

Commented by Rasheed.Sindhi last updated on 27/Dec/21

∨ ∩i⊂∈ sir! If your ′stupid′ method is so wonderful what your ′wonderful′ method will be!!!

$$\vee\:\cap\mathrm{i}\subset\in\:\boldsymbol{\mathrm{sir}}!\: \\ $$$${If}\:{your}\:'{stupid}'\:{method}\:{is}\:{so}\:{wonderful} \\ $$$${what}\:{your}\:'{wonderful}'\:{method}\:{will}\:{be}!!! \\ $$

Commented by Rasheed.Sindhi last updated on 27/Dec/21

The following method also works well: x≡y(mod 6)_(x & y have_(same remainder_(on dividing 6) ) ) ⇒10^x ≡10^y (mod 105)_(10^x &10^y have_(same remainder_(on dividing 105) ) ) ⇒10^(6x+k) ≡10^k 10^(100) =10^(6×16+4) ≡10^4 ≡25(mod 105)

$$\mathcal{T}{he}\:{following}\:{method}\:{also}\:{works}\:{well}: \\ $$$$\underset{\underset{\underset{{on}\:{dividing}\:\mathrm{6}} {{same}\:{remainder}}} {{x}\:\&\:{y}\:{have}}\:} {{x}\equiv{y}\left({mod}\:\mathrm{6}\right)}\Rightarrow\underset{\underset{\underset{{on}\:{dividing}\:\mathrm{105}} {{same}\:{remainder}}} {\mathrm{10}^{{x}} \&\mathrm{10}^{{y}} \:{have}}} {\mathrm{10}^{{x}} \equiv\mathrm{10}^{{y}} \left({mod}\:\mathrm{105}\right)} \\ $$$$\Rightarrow\mathrm{10}^{\mathrm{6}{x}+{k}} \equiv\mathrm{10}^{{k}} \\ $$$$\:\:\:\:\mathrm{10}^{\mathrm{100}} =\mathrm{10}^{\mathrm{6}×\mathrm{16}+\mathrm{4}} \equiv\mathrm{10}^{\mathrm{4}} \equiv\mathrm{25}\left({mod}\:\mathrm{105}\right) \\ $$

Commented by mr W last updated on 27/Dec/21

this is a smart method!

$${this}\:{is}\:{a}\:{smart}\:{method}! \\ $$

Commented by Rasheed.Sindhi last updated on 27/Dec/21

Thanks sir!

$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\boldsymbol{\mathrm{sir}}! \\ $$

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