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

[karush]

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

 

[kaliprasad]

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
$1+1= 2$

 

[skeeter]

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

 

[karush]

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!

 

[HallsofIvy]

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