To find the sum of two-digit numbers in an 8086 assembly program, you would typically load the two numbers into registers, add them using the ADD instruction, and store or display the result. Here's a simplified outline of the program:
MOV AX, 12h ; Load first two-digit number (18 in decimal)
MOV BX, 34h ; Load second two-digit number (52 in decimal)
ADD AX, BX ; Add the two numbers
; AX now contains the sum (70 in decimal)
This program assumes that the numbers are already defined and uses hexadecimal notation for clarity. The result can be further processed or displayed as needed.
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One many find this answer on YouTube. One also may find out how to write ascending order programs using an 8086 microprocessor by looking at the owners manual.
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
906609=913*993 I found this using a relatively simple computer program that I wrote.
There are 9000 4 digit numbers Find 4 digit numbers that are all different from 1 find 4 digit numbers that are all different from 1 The thousands digit has 8 ways to choose, hundreds, tens, and units all have 9 ways => there are 8 x 9 x 9 x 9 = 5832 digits => 4 digit numbers contain at least 1 digit 1 = number of 4 digit numbers , the number of 4 digit numbers is different by digit 1 =9000 - 5832 = 3168 numbers
Okey-dokey. I have no idea what the heck you are asking. 'What numbers do you find in the ones digit for the number 2?' Well, the number 2 is only one digit, so the answer would be, '2.' DUH!
5 and 0
To find the total number of 8-digit numbers, we consider the range of 8-digit numbers, which starts from 10,000,000 and ends at 99,999,999. The first digit (most significant) can be any digit from 1 to 9 (9 options), while the remaining 7 digits can each be any digit from 0 to 9 (10 options each). Thus, the total number of 8-digit numbers is calculated as (9 \times 10^7), which equals 90,000,000.
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program to find maximum of two numbers using pointers
20 101 a three digit number and i don't know what is the answer
Three-digit numbers range from 100 to 999. To find the count of three-digit numbers between 99 and 3000, we consider the range from 100 to 999, which includes 900 numbers (999 - 100 + 1 = 900). Thus, there are 900 three-digit numbers between 99 and 3000.