The digital product P(N) of an integer N is the product of its decimal digits.
For example, P(128)=16.
Find all sets of three positive integers A, B and C such that:
A = P(B)+P(C)
B = P(A)+P(C)
C = P(A)+P(B).
(May very well not be a complete solution; so findings)
I found:
(36,36,36) and
(498, 498, 576)
I did not find 5,5,10 because pod(10) = 0
[Corrected after Charlie's next post: (5,5,10) is indeed a solution, I made the initial blunder of eliminating all numbers that included a '0', I have commented out those lines of code below]
Part One: all equal
First, try A=B=C. Summing the 3 equations then shows A = 2*P(A).
A quick search up to 1000, for i == 2*pod(i) shows 36 with pod=18 (and 0, 0).
Proof that no number larger than 2 digits has this property:
Suppose a 3-digit num has n = 2*pod(n) (call a,b,c its digits)
100a + 10b + c = 2abc
100a + 10b = (2ab - 1)c ; c must be even since (2ab - 1) is odd.
10(10a + b) = odd * c ; c must end in zero, but it's only one digit
so c=0 but then pod(abc) = 0.
So for n more than 2 digits, the pod is zero, so no such n exists
Thus (36,36,36) is the only solution where A=B=C.
Part Two: two equal
Subtract one equation from another:
A - B = P(B) - P(A) (true for all pairs of ABC)
Find pairs of integers s.t. A - B = P(B) - P(A)
Once you find such a pair, it is possible that A,A,B or A,B,B is a solution. If true then the 2 to 1 weighted average of A,B must be double the 2 to 1 weighted average of pod(A),pod(B); or just check if A+A+B = 2(P(A)P(A)P(B))
(and also check for ABB)
The second part of the code prints out:
498 576 288 210 78
[[498, 498, 576]]
Part Three: all unique
I then started an analytic approach for if A,B,C are all unique; but no progress so far.
-----------------
from itertools import combinations_with_replacement
def pod(n):
""" Input an integer. Returns the Product of the Digits """
aList = list(str(n))
ans = 1
for c in aList:
ans = ans * int(c)
return ans
for i in range(1,1000):
if i == 2*pod(i):
print(i,pod(i))
big = 10000
ans = []
for i in range(big):
# if '0' in str(i): #commented out, see correction above
# continue #commented out
for j in range(i+1,big):
# if '0' in str(j): #commented out
# continue #commented out
if i-j == pod(j) - pod(i):
# print(i,j, pod(i), pod(j), ' ', j-i)
sum1 = i+i+j
sum2 = i+j+j
if i+i+j == 2*(pod(i)+pod(i)+pod(j)):
ans.append([i,i,j])
print(i,j, pod(i), pod(j), ' ', j-i)
if i+j+j == 2*(pod(i)+pod(j)+pod(j)):
ans.append([i,j,j])
print(i,j, pod(i), pod(j), ' ', j-i)
print(ans)
Edited on September 5, 2023, 10:54 am
Edited on September 6, 2023, 5:37 am
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Posted by Larry
on 2023-09-05 10:52:53 |
Computer solution
