An Exponential Diophantine Equation | Number Theory

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[@wtspman]

My thought was to convert 42 to binary = 101010. There are ones in three of the binary digits => those give you the values of a, b, and c. If there were ones in anything other than 3 of the binary digits then there would be no solution for the sum of three powers of 2.

[@wafflaaar1067]

woah.. this is a high IQ solution.

[@giorgiocanal1659]

Simply awesome!

[@WahranRai]

Similar to binary conversion, successive subtraction of the nearest power of 2
42 - 2^5 (32) = 10 ➔
10 - 2^3 (8) = 2 ➔
2 - 2 ^1 (2) = 0 ➔ 42 = 32 + 8 + 2 = 2^5 + 2^3 + 2^1 ➔ {a,b,c} = {5,3,1}

[@Kounomura]

The task can be completed by representing 42 in the number system of two. 42= 32+8+2=2^5+2^3+2^1 {a,b,c}={5,3,1}

[@wafflaaar1067]

How I tried doing it:
as 42 is a multiple of 2, I thought to myself:
42 = 2 * something
in fact, we can factor out a 2 from the sum of powers of 2.
we go from
2^a + 2^b + 2^c = 42
2 ( 2^(a-1) + 2^(b-1) + 2^(c-1) ) = 42
divide both sides by 2
2^(a-1) + 2^(b-1) + 2^(c-1) = 21
we have 21 on RHS and it is odd... but we have powers of 2. only odd power of 2 is 1, therefore let 2*(c-1) = 1 , c - 1 = 0, c = 1
giving...
2^(a-1) + 2^(b-1) + 1 = 21
subtract 1 frm both sides -> 2^(a-1) + 2^(b-1) = 20
now, which powers of 2 give us 20?
16 and 4 (the others... well, they dont. 8 needs 12 to make 20 but 12 is not a power of 2, 2 needs 18 to make 20, but 18 not a power of 2)
let 16 = 2^(a-1), 4 = 2^(b-1)
therefore a = 5, b = 3, c = 1

[@jaimeduncan6167]

I did it almost like you, once you have the 20 you can factor 20 as 2^2*5 divided by 2^2 and apply one more time the first step because again, 5 is odd. Then you get the response.I prefer it like that because it's more systemic, like, beautiful.

[@littlegamer944]

I have a question please solve this -
2520 = (x+y+xy)^2+2xy+2y-3x
Find possible value of y if x and y are pos integers

[@prasadbmvs]

Nice explanation

[@nasrullahhusnan2289]

(2^a)+(2^b)+(2^c)=42
=32+8+2
=3⁶+2³+2¹
(a,b,c) is permutation of 1, 3, and 6

[@scottleung9587]

Got it!

[@FisicTrapella]

2^a(1 + 2^(b-a) + 2^(c-a)) = 2·3·7
So, a=1 and 2^(b-1) + 2^(c-1) = 20
That is, you need 2 powers of 2 whose sum equals 20 and the only case is 16+4. So, b=5 and c=3.

[@broytingaravsol]

easy

[@RealQuInnMallory]

{A=1 B=20 c=21} {A=1:b=2c=3 }

[@rakenzarnsworld2]

a = 5, b = 3, c = 1

[@stevemonkey6666]

😮‍💨1,3,5 works

[@phill3986]

👍👍👍