#when is this divisible by 3?

82 messages · Page 1 of 1 (latest)

lone sparrowBOT
safe cloak
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notice that 3 divides 2^(a+2b)-2^a - 3, if and only if, 3 divides 2^(a+2b)-2^a

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simplifying this, 2^(a+2b)-2^a = 2^a(4^b-1)

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3 doesnt divide 2^a, but taking a look to 4^b -1

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we notice that the remainder of 4 when it is divided by 3 is 1

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so it happens the same with 4^b

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as we have 4^b - 1, then it is a multiple of 3

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which means that 2^(a+2b)-2^a - 3 is a multiple of 3

fickle oak
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wait wait

fickle oak
safe cloak
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sure

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as a consequence of the properties of the exponential function, we have:

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2^(a+2b) = 2^a * 2^(2b)

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and 2^(2b) = (2^2)^b = 4^b

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now, by the distributive property, we have:

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2^a * 4^b - 2^a = 2^a(4^b - 1)

manic crest
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no one needs to know...

safe cloak
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its ok, cor

manic crest
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never

safe cloak
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notice how you divided it all by 2^a and 3 isnt divisible by that, you were working with rational division, and we are talking about integer one

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but its good to remark to learn

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btw, bribe

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we have seen that 3 divides it for all a and b naturals

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if we divide by 3, we have

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2^a(4^b - 1)/3 - 1

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if its divisible by 9 (the original number), this would have to be divisible by 3, obviously

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so

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if 3 divides the whole thing

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as it doesnt divide -1, it cant divide 2^a(4^b - 1)/3

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2^a(4^b - 1)/3 would need to be 1 mod 3

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maybe studying cases you reach the result

manic crest
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no the rational division

safe cloak
manic crest
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yep

safe cloak
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that isnt an integer unless a = 0

manic crest
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yep

safe cloak
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so it didnt make sense here was we are working with integer division

manic crest
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oh ye if there wasnt a 3 at the botton it would have worked

safe cloak
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under the perspective of rational numbers, each rationaal number distinct of 0 has inverse regards the product, it is a field, so you can divide by any number distinct of 0 there

manic crest
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um bruh...

safe cloak
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more clearly, when you took the common factor 2^a

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as we are in (Z, +, *), and 2^a isnt a divisor of 3, you cant extract it from the whole expression

manic crest
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ye thats the whole point

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so the end result would be a decimal

safe cloak
manic crest
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but i didnt see the 3 as the denominator

safe cloak
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{2^a mod 3 : a is natural is finite}

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in fact, it follows a cycle

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2 mod 3 is 2
2^2 mod 3 is 1
2^3 mod 3 is 2
2^4 mod 3 is 1

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etc

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and 4^b mod 3 is 1 for all natural b

manic crest
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are you just saying what you have proved but more rigorous

safe cloak
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the rest is result of basic properties of algebra

safe cloak
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split cases for a is even and a is odd

manic crest
safe cloak
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imagine we have a + b

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and its divided by c

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imagine b isnt divided by c but a was

manic crest
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oh is this after its been divided by 3?

safe cloak
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then a + b = m * c and a = n c so:
m * c = n * c + b
and b = c(m-n)

manic crest
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we are having a mathematical discussion

lone sparrowBOT
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Solved

Post marked as solved by @dusty cedar.

Use .unsolved if this was a mistake.

safe cloak
manic crest
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this is not abstract algebruh

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did the bot refuse to make it as solved LOL

safe cloak
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xDD

safe cloak
manic crest
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ok ill look at it tmrrw when im less tired

safe cloak
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@dusty cedar

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study it but, if im correct, if a is odd, b needs to be 3 * n + 2 with n natural
and if a is even, b needs to be 3 * n + 1 with n natural

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yw

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what i did was, after proving that it was divisible by 3, i divided it by 3, then i worked with mod 3 again and by cases, taking advantage that despite there are 2 factors with powers, their modulo follows a cyclical sequence

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2^a(4^b - 1)/3 - 1 = 0 mod 3, so 2^a(4^b - 1)/3 = 1 mod 3

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meaning that 2^a is the inverse of (4^b - 1)/3 in Z_3

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if a is odd, 2^a = 2 mod 3

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so (4^b - 1)/3 needs to be 2 mod 3 too

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as the inverse is unique

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if a is even, 2^a = 1 mod 3
so (4^b - 1)/3 needs to be 1 mod 3 too