To solve the problem \(2 + 2^2 + 2^3 + \ldots + 2^{100}\) and determine when this sum is divisible by both 31 and 5, we first need to calculate the sum of this series.

The series is a geometric series, which can be summed up using the formula for the sum of a geometric series:

\[
S_n = a \frac{r^n - 1}{r - 1}
\]

where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

In this case:
- \(a = 2\)
- \(r = 2\)
- The last term is \(2^{100}\), hence there are \(100\) terms.

Substituting these values, we get:

\[
S = 2 \frac{2^{100} - 1}{2 - 1} = 2 (2^{100} - 1) = 2^{101} - 2
\]

Now, we need to check when \(2^{101} - 2\) is divisible by both 31 and 5.

### Divisibility by 31

We can use Fermat's Little Theorem, which states that if \(p\) is a prime and \(a\) is not divisible by \(p\), then:

\[
a^{p-1} \equiv 1 \mod p
\]

Applying this for \(p = 31\):

\[
2^{30} \equiv 1 \mod 31
\]

Now we need to find \(2^{101} \mod 31\):

\[
101 \mod 30 = 11 \quad \Rightarrow \quad 2^{101} \equiv 2^{11} \mod 31
\]

Next, we can compute \(2^{11} \mod 31\):

\[
2^1 = 2, \quad 2^2 = 4, \quad 2^3 = 8, \quad 2^4 = 16, \quad 2^5 = 32 \equiv 1 \mod 31
\]
\[
2^6 \equiv 2 \mod 31, \quad 2^7 \equiv 4 \mod 31, \quad 2^8 \equiv 8 \mod 31, \quad 2^9 \equiv 16 \mod 31, \quad 2^{10} \equiv 1 \mod 31, \quad 2^{11} \equiv 2 \mod 31
\]

Thus, \(2^{101} \equiv 2 \mod 31\):

\[
2^{101} - 2 \equiv 2 - 2 \equiv 0 \mod 31
\]

So, \(2^{101} - 2\) is divisible by 31.

### Divisibility by 5

We again apply Fermat's Little Theorem for \(p = 5\):

\[
2^4 \equiv 1 \mod 5
\]

Calculating \(2^{101} \mod 5\):

\[
101 \mod 4 = 1 \quad \Rightarrow \quad 2^{101} \equiv 2^1 \equiv 2 \mod 5
\]

Thus, we have:

\[
2^{101} - 2 \equiv 2 - 2 \equiv 0 \mod 5
\]

So, \(2^{101} - 2\) is also divisible by 5.

### Conclusion

Since \(2^{101} - 2\) is divisible by both 31 and 5, it is confirmed that:

\[
2 + 2^2 + 2^3 + \ldots + 2^{100}
\]

is divisible by both 31 and 5.