RPN: HP 15C and DM32: Integer Length and Digit Extraction

RPN: HP 15C and DM32: Integer Length and Digit Extraction (up to 99,999,999)

The following programs are utilities for positive integers:

(1) Integer Length: how many digits in a positive integer

(2) Digit Extraction: Get the nth digit of a positive integer

Note:  Do not use more than 8 digits (integers up to 99,999,999).   Beyond this, the program may return inaccurate answers.  

Thanks to J-F Garnier for pointing out the limitation of the programs.  

Integer Length

Find the number of digits of a positive integer. The integer is decimal base (base 10).

For a positive integer n, the number of the digits can be easily found by the formula:

L = int(log(x)) + 1

DM32, HP 32S, HP 32SII Code

L01 LBL L

L02 LOG

L03 IP

L04 1

L05 +

L06 RTN

HP 15C, DM15 Code

001	42, 21, 11	LBL A
002	43, 13	LOG
003	43, 44	INT
004	1	1
005	40	+
006	43, 32	RTN

The program finds the number of digits in the positive integer in the X stack.

Examples:

X = 436782; Length: 6

X = 5195008; Length: 7

X = 23156956; Length: 8

Digit Extraction

Extract the nth digit of a positive integer. Digit positions go from left to right. For example: for the integer 4582, the 1^st digit is 4, the 2^nd digit is 5, 3^rd digit is 8, and 4^th digit is 2.

Steps that this program follows:

Step 1: Find the length of the positive integer. L = int(log(x)) + 1.

Step 2: Divide X by 10^(L – n + 1). D = X / (10^(L – n + 1))

Step 3: Extract the fractional part. D = frac(D)

Step 4: Multiple the result by 10 and extract the integer part. D = int(10 * D)

Setting up the stack:

Y: integer

X: nth digit to extract

Variables Used:

DM32, HP 32S, HP 32S II	HP 15C, DM15
X	R1
N	R2
L	R3

DM32, HP 32S, HP 32SII Code

E01 LBL E

E02 STO N

E03 R↓

E04 STO X

E05 LOG

E06 IP

E07 1

E08 +

E09 STO L

E10 RCL X

E11 RCL L

E12 RCL- N

E13 1

E14 +

E15 10^x

E16 ÷

E17 FP

E18 10

E19 ×

E20 IP

E21 RTN

HP 15C, DM15 Code

001	42, 21, 12	LBL B
002	44, 2	STO 2
003	33	R↓
004	44, 1	STO 1
005	43, 13	LOG
006	43, 44	INT
007	1	1
008	40	+
009	44, 3	STO 3
010	45, 1	RCL 1
011	45, 3	RCL 3
012	45, 30, 2	RCL- 2
013	1	1
014	40	+
015	13	10^x
016	10	÷
017	42, 44	FRAC
018	1	1
019	0	0
020	20	×
021	43, 44	INT
022	43, 32	RTN

Examples

Integers	Length	2nd Digit	4th Digit	5th Digit
436782	6	3	7	8
5195008	7	1	5	0
23156956	8	3	5	6

Eddie

All original content copyright, © 2011-2025. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

All posts are 100% generated by human effort.  The author does not use AI engines and never will.

TI 30Xa Algorithms: Greatest Common Divisor

TI 30Xa Algorithms: Greatest Common Divisor

To find the greatest common divisor between two positive integers U and V:

Let U ≥ V. Let U = A * V + R where A is the quotient of U / V and R is the remainder. If R≠0, then V becomes the new U and R become the new V. The process repeats until R=0. At that point the value of V prior to the last calculation is the greatest common divisor (GCD) of U and V.

Example:

gcd(166, 78)

U = 166, V = 78

Algorithm Loop:

1. A = int(U / V)

2. R = U – V * int(U / V)

	A	R	U	V
Start	n/a	n/a	166	78
A = int(166 / 78) = 2 R = 166 – 2 * 78 = 10	2	10	78	10
A = int(78 / 10) = 7, R = 78 – 7 * 10 = 8	7	8	10	8
A = int(10 / 8) = 1 R = 10 – 1 * 8 = 2	1	2	8	2
A = int(8 / 2) = 4 R = 8 – 4 * 2 = 0	4	0 *STOP*

Procedure

1. Store the greater of the two numbers in memory register 1: [ STO ] [ 1 ].

2. Store the lesser of the two numbers in memory register 2: [ STO ] [ 2 ].

3. Divide memory register 1 by memory register 2. Store the integer part (no fractional part) in memory register 3: [ RCL ] [ 1 ] [ ÷ ] [ RCL ] [ 2 ] [ = ], (integer part) [ STO ] [ 3 ]

4. Figure the remainder and store the result in memory 3: [ RCL ] [ 1 ] [ - ] [ RCL ] [ 2 ] [ × ] [ RCL ] [ 3 ] [ = ] [ STO ] [ 3 ]

5. If the remainder is 0, stop. The GCD is stored in memory 2.

6. If the remainder is non-zero, then store memory 2 into memory 1 then memory 3 into memory 2. You need to do it in this order. [ RCL ] [ 2 ] [ STO ] [ 1 ], [ RCL ] [ 3 ] [ STO ] [ 2 ]. Go back to Step 3 and repeat.

Examples

Example 1: GCD(26, 14)

M1 = 26, M2 = 14

	M1	M2	M3
26 [ STO ] [ 1 ], 14 [ STO ] [ 2 ]	26	14
[ RCL ] [ 1 ] [ ÷ ] [ RCL ] [ 2 ] [ = ] Result: 1.857142857 1 [ STO ] [ 3 ]	26	14	1
[ RCL ] [ 1 ] [ - ] [ RCL ] [ 2 ] [ × ] [ RCL ] [ 3 ] [ = ] Result: 12 [ STO ] [ 3 ] R is not zero, so we continue.	26	14	12
[ RCL ] [ 2 ] [ STO ] [ 1 ], [ RCL ] [ 3 ] [ STO ] [ 2 ]	14	12	12
[ RCL ] [ 1 ] [ ÷ ] [ RCL ] [ 2 ] [ = ] Result: 1.166666667 1 [ STO ] [ 3 ]	14	12	1
[ RCL ] [ 1 ] [ - ] [ RCL ] [ 2 ] [ × ] [ RCL ] [ 3 ] [ = ] Result: 2 [ STO ] [ 3 ] R is not zero, so we continue.	14	12	2
[ RCL ] [ 2 ] [ STO ] [ 1 ], [ RCL ] [ 3 ] [ STO ] [ 2 ]	12	2	2
[ RCL ] [ 1 ] [ ÷ ] [ RCL ] [ 2 ] [ = ] Result: 6 6 [ STO ] [ 3 ]	12	2	6
[ RCL ] [ 1 ] [ - ] [ RCL ] [ 2 ] [ × ] [ RCL ] [ 3 ] [ = ] Result: 0 [ STO ] [ 3 ] R is zero, so we stop. GCD: [ RCL ] [ 2 ]: GCD(26, 14) = 2	12	2	0

Example 2: GCD(27, 15)

M1 = 27, M2 = 15

	M1	M2	M3
27 [ STO ] [ 1 ], 15 [ STO ] [ 2 ]	27	15
[ RCL ] [ 1 ] [ ÷ ] [ RCL ] [ 2 ] [ = ] Result: 1.8 1 [ STO ] [ 3 ]	27	15	1
[ RCL ] [ 1 ] [ - ] [ RCL ] [ 2 ] [ × ] [ RCL ] [ 3 ] [ = ] Result: 12 [ STO ] [ 3 ] R is not zero, so we continue.	27	15	12
[ RCL ] [ 2 ] [ STO ] [ 1 ], [ RCL ] [ 3 ] [ STO ] [ 2 ]	15	12	12
[ RCL ] [ 1 ] [ ÷ ] [ RCL ] [ 2 ] [ = ] Result: 1.25 1 [ STO ] [ 3 ]	15	12	1
[ RCL ] [ 1 ] [ - ] [ RCL ] [ 2 ] [ × ] [ RCL ] [ 3 ] [ = ] Result: 3 [ STO ] [ 3 ] R is not zero, so we continue.	15	12	3
[ RCL ] [ 2 ] [ STO ] [ 1 ], [ RCL ] [ 3 ] [ STO ] [ 2 ]	12	3	3
[ RCL ] [ 1 ] [ ÷ ] [ RCL ] [ 2 ] [ = ] Result: 4 4 [ STO ] [ 3 ]	12	3	4
[ RCL ] [ 1 ] [ - ] [ RCL ] [ 2 ] [ × ] [ RCL ] [ 3 ] [ = ] Result: 0 [ STO ] [ 3 ] R is zero, so we stop. GCD: [ RCL ] [ 2 ]: GCD(27, 15) = 3	12	3	0

I hope you find this useful. What I hope to do with the monthly series is to demonstrate various calculations with the TI-30Xa.

Note: For June and July 2024, I will be posting on Saturdays only. I plan to resume the Saturday-Sunday schedule in August.

Eddie

All original content copyright, © 2011-2024. Edward Shore. Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited. This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.