Euler 0017
The Problem:
WhatIf isyou were to write out a number as a word, you could then count the sumletters and generate a new number.
For example: 1 becomes one which becomes 3.
We use british convention which adds an 'and' between the hundreds and tens place. For example one hundred and forty-four.
How many letters are used in total to write out all of the digitsnumbers from 1 to 1000? We can exclude hyphens.
Considerations and Approach:
There are a lot of 2^1000?
Considerations:
With pythonuniques we canneed triviallyto calculatekeep 2^1000track andof, but then we can convert itnumbers off of a set of rules.
1 - one - 3
2 - two - 3
3 - three - 5
4 - four - 4
5 - five - 4
6 - six - 3
7 - seven - 5
8 - eight - 5
9 - nine - 4
10 - ten - 3
11 - eleven - 6
12 - twelve - 6
13 - thirteen - 8
14 - fourteen - 8
15 - fifteen - 7
16 - sixteen - 7
17 - seventeen - 9
18 - eighteen - 8
19 - nineteen - 8
20 - twenty - 6
30 - thirty - 6
40 - forty - 5
50 - fifty - 5
60 - sixty - 5
70 - seventy - 7
80 - eighty - 6
90 - ninety - 6
and - 3
thousand - 8
hundred - 7
1 through twenty have distinct numbers so we will need to akeep stringthat andinto sum its digits.
The Approach:
We calculate 2^1000 and then we can convert it to a string and sum its digits.account.
The Code:
powernumber_dict = {1:3,2:3,3:5,4:4,5:4,6:3,7:5,8:5,9:4,10:3,11:6,12:6,13:8,14:8,15:7,16:7,17:9,18:8,19:8,20:6,30:6,40:5,50:5,60:5,70:7,80:6,90:6}
number_end = 1000
number = 2**power
total_sum = 0
for digiti in str(number)range(1,number_end + 1):
construction = ""
hundred = False
if int(i / 1000) > 0:
total_sum += number_dict[int(digit)i / 1000)] #thousands place number
#print(i,int(i / 1000))
total_sum += 8 #the word thousand
if int((i % 1000) / 100) > 0:
#print(i,int((i % 1000) / 100))
total_sum += number_dict[int((i % 1000) / 100)] #hundreds place number
total_sum += 7 #the word hundred
hundred = True
construction += str(int((i % 1000) / 100)) + "hundred"
if int((i % 100)) > 0:
tens = int((i % 100)/10)*10
ones = int((i % 100))
ones_true = int(i%10)
if hundred:
construction += " and "
total_sum += 3 #the word and
if tens >= 20: #numbers greater than 20
construction += str(tens)
total_sum += number_dict[tens]
if ones > 9 and tens < 20: #10 through 19
construction += str(ones)
total_sum += number_dict[ones]
if ones_true > 0 and (ones < 10 or ones > 19): #ten through 19 are weird
construction += str(ones_true)
total_sum += number_dict[ones_true]
#print(construction)
print(total_sum)