I have to write a code where if I put a 3 digit number in, it will count how many hundreds, tens and ones are in it e.g 327=3 hunders, 2 tens. 7. ones

Question

num = int(input('Give me a 3 digit number! :'))

div1 = 100
div2 = 110
div3 = 36

hundreds = num // div1
tens = num // div2
ones = num // div3

print("In %d there are %d hundred(s) %d ten(s) and %d ones are found" % (num, hundreds, tens, ones))

Output (It should be; "In 187 there are 1 hundred(s) 8 ten(s) and 7 ones are found")

The real results

Give me a 3 digit number! : 187
In 187 there are 1 hundred(s) 1 ten(s) and 5 ones are found

Asked By: Shoze

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Source

Answer 1

Use the modulus operator:

num = int(input('Give me a 3 digit number! :'))

hundreds = num // 100
tens = num % 100 // 10
ones = num % 10

print("In %d there are %d hundred(s) %d ten(s) and %d ones are found" % (num, hundreds, tens, ones))

Result:

Give me a 3 digit number! :187
In 187 there are 1 hundred(s) 8 ten(s) and 7 ones are found

Answered By: CryptoFool

Answer 2

you could simply work with the string, there is no need of calculations, only index the string by order

num = input('Give me a 3 digit number! :')
assert num.isnumeric() and len(num) == 3 # this line verify that the input is a integer and has 3 digits
print(f"In {num} there are {num[0]} hundred(s) {num[1]} ten(s) and {num[2]} ones are found")

Give me a 3 digit number! :456
In 456 there are 4 hundred(s) 5 ten(s) and 6 ones are found

Answered By: Lucas M. Uriarte

Answer 3

I found myself bored while waiting for a delivery. I passed the time building a general solution to this problem:

labels = [(10, 'one'), (10, 'ten'),
          (10, 'hundred'), (1000, 'thousand'),
          (1000, 'million'), (1000, 'billion')]

def write_it_out(num):
    p = 0
    parts = []
    while num > 0 and p < len(labels):
        num, m = divmod(num, labels[p][0])
        parts.append((m, labels[p][1]))
        p += 1

    r = ""
    for part in parts[-1::-1]:
        if not part[0]:
            continue
        if r:
            r += ", "
        r += f"{part[0]} {part[1]}{'s' if part[0] > 1 else ''}"
    return r

# num = int(input('Give me a 3 digit number! :'))
# print(write_it_out(num))

def test(num):
    print(f"{num}: {write_it_out(num)}")

test(187)

test(1)
test(2)
test(440)
test(19)
test(121)
test(121003)
test(1234)
test(123456789)
test(12345678900)
test(1000000000)

Result:

187: 1 hundred, 8 tens, 7 ones
1: 1 one
2: 2 ones
440: 4 hundreds, 4 tens
19: 1 ten, 9 ones
121: 1 hundred, 2 tens, 1 one
121003: 121 thousands, 3 ones
1234: 1 thousand, 2 hundreds, 3 tens, 4 ones
123456789: 123 millions, 456 thousands, 7 hundreds, 8 tens, 9 ones
12345678900: 12 billions, 345 millions, 678 thousands, 9 hundreds
1000000000: 1 billion

It will mess up if you give it a negative number or a number of 1000 billion or greater. Fixing that by adding more labels is left to the reader.

PS: Thanks @Sterling for pointing me to divmod!

Answered By: CryptoFool

I have to write a code where if I put a 3 digit number in, it will count how many hundreds, tens and ones are in it e.g 327=3 hunders, 2 tens. 7. ones

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