MHB Adding in Base 20: 1HE1C +JDF0 = 2H7GC

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Base
AI Thread Summary
The discussion focuses on adding two numbers in base 20: 1HE1C and JDF0, resulting in 2H7GC. The calculation involves converting each digit to base 10 for easier addition, then converting back to base 20. The key steps include adding C and 0, F and 1, and E and D, which requires carrying over values when sums exceed 20. The final carry from adding H and J leads to the correct result of 2H7GC. This method emphasizes the importance of understanding base conversions for accurate arithmetic in non-decimal systems.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
Add in base 20

1HE1C +JDF0= 2H7GC from calculator
By hand I couldn't get the H

Using
1
2
3
4
5
6
7
8
9
A=10
B=11
C=12
D=13
E=14
F=15
G=16
H=17
I=18
J=19
K=20
 
Mathematics news on Phys.org
as we are not comfortable with base 20 let us convert to base 10 (digit by digit) and concert it back
1HE1C
JDF 0
--------
2H7GC

C+ 0 = C OK
F + 1 = G OK
$E + D = (14 + 13)_{10} = 27_{10} = 17_{20}$ so 7 and carry 1
$H+J + 1 = (17 + 19 + 1)_{10} = 37_{10} = {1H}_{20}$ so H and carry 1
$1+1= 2$
 
learned this method in a CS FORTRAN class a long time ago ...

Let $x=20$

1HE1C = $x^4 + 17x^3 + 14x^2 + x + 12x^0$

JDF0 = $19x^3 + 13x^2 + 15x + 0x^0$

sum ...

$x^4 + (20+16)x^3 + (20+7)x^2 + 16x + 12x^0$

$x^4 + 20x^3 + 16x^3 + 20x^2 + 7x^2 + 16x + 12x^0$

$x^4 + x^4 + 16x^3 + x^3 + 7x^2 + 16x + 12x^0$

$2x^4 + 17x^3 + 7x^2 + 16x + 12x^0$ = 2H7GC
 
skeeter said:
learned this method in a CS FORTRAN class a long time ago ...

Let $x=20$

1HE1C = $x^4 + 17x^3 + 14x^2 + x + 12x^0$

JDF0 = $19x^3 + 13x^2 + 15x + 0x^0$

sum ...

$x^4 + (20+16)x^3 + (20+7)x^2 + 16x + 12x^0$

$x^4 + 20x^3 + 16x^3 + 20x^2 + 7x^2 + 16x + 12x^0$

$x^4 + x^4 + 16x^3 + x^3 + 7x^2 + 16x + 12x^0$

$2x^4 + 17x^3 + 7x^2 + 16x + 12x^0$ = 2H7GC

that would be a little bit easier to remember!
 
karush said:
Add in base 20

1HE1C +JDF0= 2H7GC from calculator
By hand I couldn't get the H
First, C+ 0= C. That's the right-most "digit".
1+ F= 1+ 15= 16= G. That's the next "digit".
E+ D= 14+ 13= 27= 20+ 7= 17. The next "digit" is 7 and "carry the 1".
That may be what's keeping you from "getting the H".
H+ J= 17+ 19= 36 and "carrying the 1", 27= 20+ 17= 1H. The next "digit" is H and "carry the 1"
With that carry, the last addition is 1+ 1= 2.
That's how we get 2H7GC!

Using
1
2
3
4
5
6
7
8
9
A=10
B=11
C=12
D=13
E=14
F=15
G=16
H=17
I=18
J=19
K=20
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top