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How to solve the Project Euler 32 problem

Armaan Nougai

Problem statement

The product 7254 is interesting because the identity, 39 × 186 = 7254, is 1 through 9 or pandigital. All three multipliers, multiplicand, and product together are pandigital (have all 1-9 digits) exactly once.

Find the sum of all these products whose multiplicand-multiplier-product-identity are pandigital (1-9).

Solution approach

Let:

 Identity --> a * b = c
 Multiplicand = a
 Multiplier = b
 Product = c
 condition  = digits of a,b,c all together makes 1-9 pandigital.

To look for every possible combination of (a,b) for which the condition is true would work in the brute force approach as shown above.

However, we still to determine the limit to which we’ll need to look for combinations of (a,b).

Finding the upper limit

Here are a few observations:

  • The number of digits of a,b,c combined must be equal to 10.

  • The term multiplicand (a) and the multiplier (b) are replaceable, i.e., a * b = c and b * a = c. So, while looking for desired combinations of multipliers and multiplicand, both (a, b) and (b, a) will satisfy the condition. However, we have to count them only once as they are practically the same.

Table

The columns represent the number of digits in b in the tables below. The rows show the number of digits in a, and the value in each cell represents the possible number of digits of c.

In the table below, there are four highlighted cases. However, this is because the multiplier and multiplicand are replaceable. Thus, we will consider the purple-highlighted cases as two and also as the only possible cases in which the number of digits of a,b,c combined could be equal to 10.

  • The first case:

    • a is double digit, b is triple digit, and c should be 5-digit.

    • The maximum possible double digit number is: a = 98 and the maximum possible 3-digit number is: b = 987.

    • Thus, the upper bound will be a=98 and b=987.

  • The second case:

    • a is a single digit, b is 4-digit, and c should be 5-digit.

    • The maximum possible single digit is a = 9. The maximum possible 4-digit b = 9876.

    • Thus, the upper bound will be a=9 and b=9876.

Implementation

We only have two possible cases. We’ll run a separate for-loop for each case while counting the sum of all the products (c), which satisfies the condition.



b is 1-digit

b is 2-digit

b is 3-digit

b is 4-digit

a is 1-digit

2-digit

2-3 digit

3-4 digit

4-5 digit

a is 2-digit

2-3 digit

3-4 digit

4-5 digit

5-6 digit

a is 3-digit

3-4 digit

4-5 digit

5-6 digit

6-7 digit

a is 4-digit

4-5 digit

5-6 digit

6-7 digit

7-6 digit

sumC = 0
alreadyFound=[]

# CASE 1. (a - 2-digit) and (b - 3-digit) 
upperA = 98
upperB = 987
for a in range(9,upperA+1):
    for b in range(98,upperB+1):
        c = a*b
        if len(str(c))>5: break
        
        allDigits = sorted(str(c)+str(a)+str(b))
    
        if allDigits == list('123456789') :
            if c not in alreadyFound :
                alreadyFound+=[c]
                sumC+=c


# CASE 2. (a - 1-digit) and (b - 4-digit) 
upperA = 9
upperB = 9876
for a in range(1,upperA+1):
    for b in range(987,upperB+1):

        c = a*b
        if len(str(c))>5: break
        
        allDigits = sorted(str(c)+str(a)+str(b))
    
        if allDigits == list('123456789') :
            if c not in alreadyFound :
                alreadyFound+=[c]
                sumC+=c


print(sumC)

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project euler
32
pandigital
communitycreator
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