The 4-digit positive integer 2022 has an interesting property. It has precisely three equal digits and is divisible by the sum of the digits, that is, 6.
Now the quotient obtained by dividing this number by the sum of the digits is: (2022)/6=337, which is a 3-digit number with precisely two equal digits.
Determine six positive 4-digit positive integers, each of them having the above-mentioned properties, that immediately follows 2022.
How many 4-digit positive integers (no leading zeros) with the foregoing properties precede 2022?
*** Remember: the quotient must have precisely 3 digits, exactly two of which must be equal.
There are a total of 13 such numbers,
one is smaller than 2022 and 11 are larger.
The six that immediately follow 2022 are marked with (*)
N N/sod(N)
1011 337
2022 337
3033 337 *
3383 199 *
3888 144 *
4044 337 *
4454 262 *
5055 337 *
6066 337
7077 337
7776 288
8088 337
9099 337
Total: 13
Before 2022: 1
After 2022: 11
------------------- code -------------------
def sod(n):
""" Input an integer. Returns the Sum of the Digits """
aList = list(str(n))
ans = 0
for c in aList:
ans = ans + int(c)
return ans
ct = 0
ctbefore = 0
ctafter = 0
it_is_before = True
it_is_2022 = False
it_is_after = False
for n in range(1000,10000):
s = str(n)
if len(set(s)) != 2:
continue
char = s[0]
count = 0
for c in s:
if c == char:
count += 1
if count == 2:
continue
summa = sod(n)
if n%summa != 0:
continue
quo = int(n/summa)
if len(str(quo)) != 3 or len(set(str(quo))) != 2:
continue
print(n, int(n/summa))
if n == 2022:
it_is_before = False
it_is_2022 = True
it_is_after = True
if it_is_before:
ctbefore += 1
elif it_is_after and not it_is_2022:
ctafter += 1
if it_is_2022:
it_is_2022 = False
ct += 1
print('Total:',ct)
print('Before 2022:',ctbefore, '\nAfter 2022:',ctafter)
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Posted by Larry
on 2022-04-26 08:16:32 |
Solution
