What is the flowchart to determine if a number is an Armstrong number?

Answer 1

Step 1: Start

Step 2: Read n

Step 3: Temp=n

Step 4: arm=0

Step 5: If n is not equal to 0,

rem=n%0

arm=arm+(rem*rem*rem)

Step 6: n=n/10, goto 5

Step 7: If n=0, then if temp=arm, print No. is armstrong

Step 8: if temp is not equal to arm print No. is not armstrong

Step 9: Stop

Answer 2

1. start the program (Connector Symbol- Ovel Shape)

2. Get the input (a number) from the user (n) - (Input/Output Symbol-parallelogram)

3. store the number in an int variable (m=n)

4. Initialize the sum as Zero (sum=0)-(Rectangle)

5. check the decision as while(n>0)-(Diamond Shape)conditional (or decision), represented as a diamond (rhombus)

6.If this yields true then process as follows:(Processing Symbols-Rectangles)

(remainder) r=n%10;

sum=sum+(r*r*r);

n=n/10;

7.check the decision statement again and continue with the process until the condition becomes false.

8.Finally check m with sum

9. If the sum==m Then print the number is an Armstrong-(Output Symbol-parallelogram)

10.Else not an Armstrong

11.Terminate the Program.(Ovel shape)

Read more: Flow_chart_to_find_the_given_number_is_an_Armstrong_number_or_not

Answer 3

We cannot show flowcharts in any meaningful way in a text-only forum such as this. However, we can demonstrate the algorithm using pseudocode.

To determine if a given number is an Armstrong number we first need to count the digits. This is achieved with the following algorithm:

Algorithm: count_digits Is:

Input: a positive integer, n

Output: the count of digits in n

count := 0 // initialise the count

repeat // iterate at least once (there has to be at least one digit)

{

count := count + 1 // increment the count

n := n / 10 // right-shift all digits one decimal place (lose the least-significant digit)

}

until n = 0 // exit the loop when there are no more digits

return count

We then need to raise each digit by the power of the count. For that we use the following algorithm:

Algorithm: power Is:

Input: two positive integers, n and p

Output: the value of n raised to the power of p

accumulator := n

while p > 1 do

{

accumulator = accumulator * n

p := p - 1

}

return accumulator

With these two algorithms we can now determine if a number is an Armstrong number or not:

Algorithm: is_armstrong

Input: a positive integer, n

Output: true if n is an Armstrong number, otherwise false

tmp := n // store the number (we'll need this later to check the result)

count := count_digits (n)

sum := 0 // initialise a sum

while n > 0 do

{

digit := n % 10 // calculate the least-significant digit (% is the modulo operator)

sum := sum + power (digit, count) // add on the digit raised to the power of count

n := n / 10 // shift all digits one position to the right

}

end while

return sum = tmp // return true if the sum is the same as the original number, otherwise false