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What is the multiplicative inverse of 4^67 (mod 19)?
From [math]19 \mid (2^{18}-1)=(2^9–1)(2^9+1)[/math], we know that [math]19 \mid (2^9–1)[/math] or [math]19 \mid (2^9+1)[/math]. We verify that [math]19 \mid (2^9+1)[/math].Now [math]4^{67} = 2^{134} = \big(2^9\big)^{15} \cdot 2^{-1} \equiv -2^{-1} \equiv -10 \equiv 9\pmod{19}[/math],and since [math]2 \cdot 9 \equiv -1\pmod{19}[/math], [math]9^{-1} \equiv -2 \equiv 17\pmod{19}[/math].Thus, [math]\big(4^{67}\big)^{-1} \equiv 17\pmod{19}[/math]. [math]\blacksquare[/math]
What's the remainder of [math]\frac{11^{99}+12^{99}+13^{99}+14^{99}+15^{99}}{196}[/math]?
This is answer avoids any computations that cannot be done in one’s head or relatively quickly on paper.First, it is crucial to recognize that [math]196[/math] is [math]14^2[/math]. This tells us that we can eliminate [math]14^{99}[/math] from our calculations, as [math]14^2[/math] divides into it, so it thus contributes no remainder.We then want to rewrite the expression as follows:[math](14-3)^{99}+(14-2)^{99}+(14-1)^{99}+(14+1)^{99} \[/math][math]mod[/math][math] 14^2[/math]This sets us up nicely to use the Binomial Theorem, which states:[math]\displaystyle (x+y)^n = \sum_{k=0}^{n} {{n}\choose{k}}x^ky^{n-k}[/math]In our case, if we were to expand each of the binomials of degree [math]99[/math] above, we realize that [math]14^2[/math] divides into every term that has a factor of [math]14^n, n\geq 2[/math]. Thus, we can eliminate every term from each binomial except for two:[math](14+x)^{99} \[/math][math]mod[/math][math] 14^2 [/math][math][/math][math][/math][math]\equiv {{99}\choose{1}}14^1x^{98}+{{99}\choose{0}}14^0x^{99} \[/math][math]mod[/math][math] 14^2[/math][math]= 99\cdot14x^{98}+x^{99} \[/math][math]mod[/math][math] 14^2[/math]Applying [math]x=-3,-2,-1,[/math] and [math]1[/math] leaves us with:[math]99\cdot14\cdot3^{98}\,-\,3^{99}\,+\,99\cdot14\cdot2^{98}\,-\,2^{99}+99\cdot14-1+99\cdot14+1 \[/math][math]mod[/math][math] 196[/math][math]99\cdot14[/math] is easy enough to calculate; it’s [math]100\cdot14-14=1386[/math]. So we have:[math]1386\cdot3^{98} - 3^{99} [/math][math][/math][math]+ 1386\cdot2^{98} - 2^{99} [/math][math][/math][math]+ 2\cdot1386 \[/math][math]mod[/math][math] 196[/math][math]= 462\cdot3^{99} - 3^{99} [/math][math][/math][math]+ 693\cdot2^{99} - 2^{99} [/math][math][/math][math]+ 2\cdot1386 \[/math][math]mod[/math][math] 196[/math][math]= 461\cdot3^{99} [/math][math][/math][math]+ 692\cdot2^{99} [/math][math][/math][math]+ 2\cdot1386 \[/math][math]mod[/math][math] 196[/math]Now we just need to find the remainder for each of the above terms.For term #1:[math]461\cdot3^{99} \[/math][math]mod[/math][math] 196[/math]Let’s look at [math]3^{99} \[/math][math]mod[/math][math] 196[/math]. It is helpful to split this up into (1) [math]3^{99} \[/math][math]mod[/math][math] 49[/math] and (2) [math]3^{99} \[/math][math]mod[/math][math] 4[/math]. (We must split [math]196[/math] into factors such that one factor is not a factor of the other factor; e.g. [math]28[/math] and [math]7[/math] may not work.)(1) [math]\pmod{49}[/math]:[math]3^{99} [/math][math][/math][math][/math][math]= 81\cdot3^{95} [/math][math][/math][math][/math][math]= 81\cdot243^{19} [/math][math][/math][math][/math][math]\equiv 81\cdot(-2)^{19} [/math][math][/math][math][/math][math]\equiv 32\cdot(-2)^{19} [/math][math][/math][math][/math][math]= -2^{24} [/math][math][/math][math][/math][math]= -256^3 [/math][math][/math][math][/math][math]\equiv -11^3 [/math][math][/math][math][/math][math]= -11\cdot121 [/math][math][/math][math][/math][math]\equiv -11\cdot23 [/math][math][/math][math][/math][math]= -253 [/math][math][/math][math][/math][math]\equiv -8 \equiv 41 \pmod{49}[/math](2) [math]\pmod{4}[/math]:[math]3^{99} [/math][math][/math][math][/math][math]\equiv (-1)^{99} [/math][math][/math][math][/math][math]\equiv 3 \pmod{4}[/math] (Much shorter!)So we need a number that has a remainder of [math]41[/math] when divided by [math]49[/math] and a remainder of [math]3[/math] when divided by [math]4[/math]. We try numbers satisfying the former until we satisfy both criteria:[math]41[/math] — no; [math]41 \equiv 1 \pmod{4}[/math][math]90[/math] — no; [math]90 \equiv 2 \pmod{4}[/math][math]139[/math] — Yes! [math]139 \equiv 3 \pmod{4}[/math]So we convert our original expression from [math]461\cdot3^{99} \[/math][math]mod[/math][math] 196[/math] to [math]461\cdot139 \[/math][math]mod[/math][math] 196[/math]. We proceed from here with [math]\[/math][math]mod[/math][math] 196[/math]:[math]461\cdot139 [/math][math][/math][math][/math][math]\equiv 69\cdot139 [/math][math][/math][math][/math][math]= 23\cdot417 [/math][math][/math][math][/math][math]\equiv 23\cdot25 [/math][math][/math][math][/math][math]= 23\cdot20+23\cdot5 [/math][math][/math][math][/math][math]= 460+115 [/math][math][/math][math][/math][math]\equiv 68+115 [/math][math][/math][math][/math][math]\equiv 183 \pmod{196}[/math]The remainder contributed by term 1 is thus [math]183[/math]. Whew. Onto term 2.For term #2:[math]692\cdot2^{99} \[/math][math]mod[/math][math] 196[/math]We look at [math]2^{99} \[/math][math]mod[/math][math] 196[/math] first. We again split this up into (1) [math]2^{99} \[/math][math]mod[/math][math] 49[/math] and (2) [math]2^{99} \[/math][math]mod[/math][math] 4[/math].(1) [math]\pmod{49}[/math]:[math]2^{99} [/math][math][/math][math][/math][math]= 8\cdot2^{96} [/math][math][/math][math][/math][math]= 8\cdot256^{12} [/math][math][/math][math][/math][math]\equiv 8\cdot11^{12} [/math][math][/math][math][/math][math]= 8\cdot121^6 [/math][math][/math][math][/math][math]\equiv 8\cdot23^6 [/math][math][/math][math][/math][math]= 8\cdot529^3 [/math][math][/math][math][/math][math]\equiv 8\cdot(-10)^3 [/math][math][/math][math][/math][math]= -800\cdot10 [/math][math][/math][math][/math][math]\equiv -16\cdot10 [/math][math][/math][math][/math][math]= -160 [/math][math][/math][math][/math][math]\equiv -13 \equiv 36 \pmod{49}[/math](2) [math]\pmod{4}[/math]:[math]2^{99} [/math][math][/math][math][/math][math]= 2\cdot2^{98} [/math][math][/math][math][/math][math]= 2\cdot4^{49} [/math][math][/math][math][/math][math]\equiv 2 \pmod{4}[/math] (Again, much shorter!)So we need a number that has a remainder of [math]36[/math] when divided by [math]49[/math] and a remainder of [math]2[/math] when divided by [math]4[/math]. We try numbers satisfying the former until we satisfy both criteria:[math]36[/math] — no; [math]36 \equiv 0 \pmod{4}[/math][math]85[/math] — no; [math]85 \equiv 1 \pmod{4}[/math][math]134[/math][math][/math] — Yes! [math]139 \equiv 2 \pmod{4}[/math]So we convert our original expression from [math]692\cdot2^{99} \[/math][math]mod[/math][math] 196[/math] to [math]692\cdot134 \[/math][math]mod[/math][math] 196[/math]. We proceed from here with [math]\[/math][math]mod[/math][math] 196[/math]:[math]692\cdot134 [/math][math][/math][math][/math][math]\equiv 104\cdot134 [/math][math][/math][math][/math][math]= 52\cdot268 [/math][math][/math][math][/math][math]\equiv 52\cdot72 [/math][math][/math][math][/math][math]= 208\cdot18 [/math][math][/math][math][/math][math]\equiv 12\cdot18 [/math][math][/math][math][/math][math]= 216 \equiv 20 \pmod{196}[/math]The remainder contributed by term 2 is thus 20.For term #3:I hope that at this point, you will be able to figure out for yourself that [math]2\cdot1386 \equiv 28 \pmod{196}[/math].Conclusion:We add all the remainders and divide by [math]196[/math].[math]183+20+28 = 231 \equiv \boxed{35 \pmod{196}}[/math]
What is an integer x from 0 to 10 such that x ≡ (134)^121 mod 11?
[math]134^{121}\equiv 2^{121} \pmod{11}[/math]Notice, [math]2^{10}=1024\equiv 1 \pmod{11}\implies 2^{121}=(2^{10})^{12}\cdot 2^1\equiv 1^{12}\cdot 2^1\equiv 2\pmod{11}[/math]
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