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Question Number 13606 by Tinkutara last updated on 21/May/17
Show that 19^(93) − 13^(99) is a positive integer divisible by 162.
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{19}^{\mathrm{93}} \:−\:\mathrm{13}^{\mathrm{99}} \:\mathrm{is}\:\mathrm{a}\:\mathrm{positive} \\ $$$$\mathrm{integer}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{162}. \\ $$
Answered by mrW1 last updated on 21/May/17
log (19^(93) )=93×log 19≈119 log (13^(99) )=99×log 13≈110 ⇒19^(93) > 13^(99) 19^(93) =(1+18)^(93) =1+93×18+((93×92)/(2!))×18^2 +((93×92×91)/(3!))×18^3 +... =A_0 +A_1 +A_2 +A_3 +...+A_(93) since 18^2 =324=2×162 ⇒A_(k≥2) mod 162 =0 A_0 +A_1 =1+93×18=1675=10×162+55 ⇒19^(93) mod 162=55 13^(99) =(1+12)^(99) =1+99×12+((99×98)/(2!))×12^2 +((99×98×97)/(3!))×12^3 +... =B_0 +B_1 +B_3 +B_4 +...+B_(99) since 12^4 =20736=128×162 ⇒B_(k≥4) mod 162 =0 3×12^3 =5184=32×162 ⇒B_3 mod 162 =0 9×12^2 =1296=8×162 ⇒B_2 mod 162 =0 B_0 +B_1 =1+99×12=1189=7×162+55 ⇒13^(99) mod 162=55 ⇒(19^(93) −13^(99) ) mod 162 =55−55=0
$$\mathrm{log}\:\left(\mathrm{19}^{\mathrm{93}} \right)=\mathrm{93}×\mathrm{log}\:\mathrm{19}\approx\mathrm{119} \\ $$$$\mathrm{log}\:\left(\mathrm{13}^{\mathrm{99}} \right)=\mathrm{99}×\mathrm{log}\:\mathrm{13}\approx\mathrm{110} \\ $$$$\Rightarrow\mathrm{19}^{\mathrm{93}} \:>\:\mathrm{13}^{\mathrm{99}} \\ $$$$ \\ $$$$\mathrm{19}^{\mathrm{93}} =\left(\mathrm{1}+\mathrm{18}\right)^{\mathrm{93}} =\mathrm{1}+\mathrm{93}×\mathrm{18}+\frac{\mathrm{93}×\mathrm{92}}{\mathrm{2}!}×\mathrm{18}^{\mathrm{2}} +\frac{\mathrm{93}×\mathrm{92}×\mathrm{91}}{\mathrm{3}!}×\mathrm{18}^{\mathrm{3}} +… \\ $$$$={A}_{\mathrm{0}} +{A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} +…+{A}_{\mathrm{93}} \\ $$$${since}\:\mathrm{18}^{\mathrm{2}} =\mathrm{324}=\mathrm{2}×\mathrm{162} \\ $$$$\Rightarrow{A}_{{k}\geqslant\mathrm{2}} \:{mod}\:\mathrm{162}\:=\mathrm{0} \\ $$$${A}_{\mathrm{0}} +{A}_{\mathrm{1}} =\mathrm{1}+\mathrm{93}×\mathrm{18}=\mathrm{1675}=\mathrm{10}×\mathrm{162}+\mathrm{55} \\ $$$$\Rightarrow\mathrm{19}^{\mathrm{93}} \:{mod}\:\mathrm{162}=\mathrm{55} \\ $$$$ \\ $$$$\mathrm{13}^{\mathrm{99}} =\left(\mathrm{1}+\mathrm{12}\right)^{\mathrm{99}} =\mathrm{1}+\mathrm{99}×\mathrm{12}+\frac{\mathrm{99}×\mathrm{98}}{\mathrm{2}!}×\mathrm{12}^{\mathrm{2}} +\frac{\mathrm{99}×\mathrm{98}×\mathrm{97}}{\mathrm{3}!}×\mathrm{12}^{\mathrm{3}} +… \\ $$$$={B}_{\mathrm{0}} +{B}_{\mathrm{1}} +{B}_{\mathrm{3}} +{B}_{\mathrm{4}} +…+{B}_{\mathrm{99}} \\ $$$${since}\:\mathrm{12}^{\mathrm{4}} =\mathrm{20736}=\mathrm{128}×\mathrm{162} \\ $$$$\Rightarrow{B}_{{k}\geqslant\mathrm{4}} \:{mod}\:\mathrm{162}\:=\mathrm{0} \\ $$$$\mathrm{3}×\mathrm{12}^{\mathrm{3}} =\mathrm{5184}=\mathrm{32}×\mathrm{162} \\ $$$$\Rightarrow{B}_{\mathrm{3}} \:{mod}\:\mathrm{162}\:=\mathrm{0} \\ $$$$\mathrm{9}×\mathrm{12}^{\mathrm{2}} =\mathrm{1296}=\mathrm{8}×\mathrm{162} \\ $$$$\Rightarrow{B}_{\mathrm{2}} \:{mod}\:\mathrm{162}\:=\mathrm{0} \\ $$$$ \\ $$$${B}_{\mathrm{0}} +{B}_{\mathrm{1}} =\mathrm{1}+\mathrm{99}×\mathrm{12}=\mathrm{1189}=\mathrm{7}×\mathrm{162}+\mathrm{55} \\ $$$$\Rightarrow\mathrm{13}^{\mathrm{99}} \:{mod}\:\mathrm{162}=\mathrm{55} \\ $$$$ \\ $$$$\Rightarrow\left(\mathrm{19}^{\mathrm{93}} −\mathrm{13}^{\mathrm{99}} \right)\:{mod}\:\mathrm{162}\:=\mathrm{55}−\mathrm{55}=\mathrm{0} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/May/17
so beautiful.
$${so}\:{beautiful}. \\ $$
Commented by RasheedSindhi last updated on 22/May/17
Great!
$$\mathrm{Great}! \\ $$
Commented by Tinkutara last updated on 22/May/17
Thanks but how do you calculated 93 log 19 and 99 log 13? In the textbook it was mentioned that use of calculators is not allowed. Second, can we use in general this property? (1 + x)^n (mod y) = (1 + nx)(mod y) I found it from internet from Quora by typing the same question.
$$\mathrm{Thanks}\:\mathrm{but}\:\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{calculated} \\ $$$$\mathrm{93}\:\mathrm{log}\:\mathrm{19}\:\mathrm{and}\:\mathrm{99}\:\mathrm{log}\:\mathrm{13}?\:\mathrm{In}\:\mathrm{the}\:\mathrm{textbook} \\ $$$$\mathrm{it}\:\mathrm{was}\:\mathrm{mentioned}\:\mathrm{that}\:\mathrm{use}\:\mathrm{of} \\ $$$$\mathrm{calculators}\:\mathrm{is}\:\mathrm{not}\:\mathrm{allowed}. \\ $$$$\mathrm{Second},\:\mathrm{can}\:\mathrm{we}\:\mathrm{use}\:\mathrm{in}\:\mathrm{general}\:\mathrm{this} \\ $$$$\mathrm{property}? \\ $$$$\left(\mathrm{1}\:+\:{x}\right)^{{n}} \left(\mathrm{mod}\:{y}\right)\:=\:\left(\mathrm{1}\:+\:{nx}\right)\left(\mathrm{mod}\:{y}\right) \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{it}\:\mathrm{from}\:\mathrm{internet}\:\mathrm{from}\:\mathrm{Quora} \\ $$$$\mathrm{by}\:\mathrm{typing}\:\mathrm{the}\:\mathrm{same}\:\mathrm{question}. \\ $$
Commented by RasheedSindhi last updated on 22/May/17
Mentioned logs can be calculated by table of logs instead of calculator.
$$\mathrm{Mentioned}\:\mathrm{logs}\:\mathrm{can}\:\mathrm{be}\:\mathrm{calculated} \\ $$$$\mathrm{by}\:\mathrm{table}\:\mathrm{of}\:\mathrm{logs}\:\mathrm{instead}\:\mathrm{of}\:\mathrm{calculator}. \\ $$$$ \\ $$
Commented by Tinkutara last updated on 22/May/17
But it was to be proved without tables. I have done it without tables as follows: (((19)/(13)))^2 = ((361)/(169)) > 2 ⇒ (((19)/(13)))^8 > 2^4 > 13 ⇒ 19^8 > 13^9 ⇒ 19^(88) > 13^(99) ⇒ 19^(93) > 13^(99) The only thing: Can we use (1 + x)^n (mod y) = (1 + nx)(mod y)? What are the conditions for x, y and n?
$$\mathrm{But}\:\mathrm{it}\:\mathrm{was}\:\mathrm{to}\:\mathrm{be}\:\mathrm{proved}\:\mathrm{without}\:\mathrm{tables}. \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{done}\:\mathrm{it}\:\mathrm{without}\:\mathrm{tables}\:\mathrm{as}\:\mathrm{follows}: \\ $$$$\left(\frac{\mathrm{19}}{\mathrm{13}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{361}}{\mathrm{169}}\:>\:\mathrm{2}\:\Rightarrow\:\left(\frac{\mathrm{19}}{\mathrm{13}}\right)^{\mathrm{8}} \:>\:\mathrm{2}^{\mathrm{4}} \:>\:\mathrm{13} \\ $$$$\Rightarrow\:\mathrm{19}^{\mathrm{8}} \:>\:\mathrm{13}^{\mathrm{9}} \\ $$$$\Rightarrow\:\mathrm{19}^{\mathrm{88}} \:>\:\mathrm{13}^{\mathrm{99}} \\ $$$$\Rightarrow\:\mathrm{19}^{\mathrm{93}} \:>\:\mathrm{13}^{\mathrm{99}} \\ $$$$\mathrm{The}\:\mathrm{only}\:\mathrm{thing}:\:\mathrm{Can}\:\mathrm{we}\:\mathrm{use} \\ $$$$\left(\mathrm{1}\:+\:{x}\right)^{{n}} \left(\mathrm{mod}\:{y}\right)\:=\:\left(\mathrm{1}\:+\:{nx}\right)\left(\mathrm{mod}\:{y}\right)? \\ $$$$\mathrm{What}\:\mathrm{are}\:\mathrm{the}\:\mathrm{conditions}\:\mathrm{for}\:{x},\:{y}\:\mathrm{and}\:{n}? \\ $$
Commented by prakash jain last updated on 22/May/17
The first step is not essential for proof. Rest of the proof remains valid even if the first step is omitted.
$$\mathrm{The}\:\mathrm{first}\:\mathrm{step}\:\mathrm{is}\:\mathrm{not}\:\mathrm{essential}\:\mathrm{for} \\ $$$$\mathrm{proof}.\: \\ $$$$\mathrm{Rest}\:\mathrm{of}\:\mathrm{the}\:\mathrm{proof}\:\mathrm{remains}\:\mathrm{valid} \\ $$$$\mathrm{even}\:\mathrm{if}\:\mathrm{the}\:\mathrm{first}\:\mathrm{step}\:\mathrm{is}\:\mathrm{omitted}. \\ $$
Commented by prakash jain last updated on 22/May/17
The expression need not be +ve for calculating mod. −11≡0≡−22 (mod 11)
$$\mathrm{The}\:\mathrm{expression}\:\mathrm{need}\:\mathrm{not}\:\mathrm{be}\:+\mathrm{ve} \\ $$$$\mathrm{for}\:\mathrm{calculating}\:\mathrm{mod}. \\ $$$$−\mathrm{11}\equiv\mathrm{0}\equiv−\mathrm{22}\:\left(\mathrm{mod}\:\mathrm{11}\right) \\ $$

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