ProbA.md

Problem A

Part 1

• There are five numbers between 1 and 500 which have the following property: if $n=d_{1} d_{2} \cdots d_{k}$ then $n=d_{1}^{3}+d_{2}^{3}+\cdots+d_{k}^{3}$ (e.g. $153=1^{3}+5^{3}+3^{3}$ ). Find these numbers.

So this isn't too difficult to implement, the part that I can't figure out is how to prove that all these numbers would be less than 500.

First take the following definitions:

$$ \begin{aligned} f(n) = \sum^n_{i=1}\left[ d_i^3 \right], \qquad \forall n \in \mathbb{Z}^+ \end{aligned} $$

Extract a list of digits

First we need to find a way to extract all the elements from a number, if this was string we might use the .split() attribute (equivalently strsplit in R), but because it's a number it will actually behave similar to a list of numbers and so instead List Comprehensions can be used.

num = 238923
digits = [ str(num)[i] for i in range(len(str(num)))]
print(digits)

['2', '3', '8', '9', '2', '3']

This gives us a list of str though, so instead, inside the comprehension we can wrap the value in int() to get something useful.

num = 238923
digits = [ int(str(num)[i]) for i in range(len(str(num)))]
print(digits)

['2', '3', '8', '9', '2', '3']

Sum the digits

Next it is necessary to define a way to return $\sum^n_{i=1}\left[d_i\right]$, in this case I've just used a loop to go through the digits.

def sumDigits(num):
    digits = [ int(str(num)[i]) for i in range(len(str(num)))]

    val = 0
    for i in digits:
        val = i**3 + val
    return val

answer = sumDigits(983291)
print(answer)

2006

In order to find when $f(n) = n$, an empty list can be initialised and for each value in the range from 1 to 500, if that property is satisfied the value can be appended to the list:

matches = []
for i in range(500):
    if sumDigits(i) == i:
        matches.append(i)

print(matches)

[0, 1, 153, 370, 371, 407]

Conclusion

So putting this all together:

def main(maxVal):
    matches = []
    for i in range(maxVal):
        if sumDigits(i) == i:
            matches.append(i)

    return matches

def sumDigits(num):
    digits = [ int(str(num)[i]) for i in range(len(str(num)))]

    val = 0
    for i in digits:
        val = i**3 + val
    return val


main(500)

[0, 1, 153, 370, 371, 407]

hence,

$$ \begin{aligned} n = \sum^n_{i=1}\left[ d_i^3 \right], \qquad \forall n \in \left{ 0, 1, 153, 370, 371, 407 \right} \end{aligned} $$

where:

• $n$ is composed of digits $d_1 d_2 d_3 ...$