February 13, 1976 was an interesting date. The product of the month, day, and 2-digit year is equal to the 4-digit year:
2*13*76=1976.
Find all dates in the 20th and 21st centuries with this property.
I found 21 such dates:
January 21, 1995: 1*21*95 = 1995
January 26, 1976: 1*26*76 = 1976
January 26, 2080: 1*26*80 = 2080
February 13, 1976: 2*13*76 = 1976
February 13, 2080: 2*13*80 = 2080
March 7, 1995: 3*7*95 = 1995
March 13, 1950: 3*13*50 = 1950
March 17, 1938: 3*17*38 = 1938
March 17, 2040: 3*17*40 = 2040
March 27, 2025: 3*27*25 = 2025
April 24, 1920: 4*24*20 = 1920
June 16, 1920: 6*16*20 = 1920
June 21, 2016: 6*21*16 = 2016
July 3, 1995: 7*3*95 = 1995
July 11, 1925: 7*11*25 = 1925
July 18, 2016: 7*18*16 = 2016
August 12, 1920: 8*12*20 = 1920
September 9, 2025: 9*9*25 = 2025
September 14, 2016: 9*14*16 = 2016
November 7, 1925: 11*7*25 = 1925
December 8, 1920: 12*8*20 = 1920
The next time this will occur is March 27, 2025.
I did not check for illegal dates such as February 31, but fortuitously, the largest day value the program found was 26.
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month_names = ['',
'January',
'February',
'March',
'April',
'May',
'June',
'July',
'August',
'September',
'October',
'November',
'December']
dates = []
for month in range(1,13):
for day in range(1,32):
prod = month * day
for year in range(1901,2100):
if year == 2000:
continue
if prod == year / (year%100):
dates.append('{} {}, {}: {}*{}*{} = {}'.format(month_names[month],
day,
year,
month,
day,
year%100,
month*day*(year%100)))
for d in dates:
print(d)
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Posted by Larry
on 2024-05-27 09:30:55 |
Computer solution
