Consider the following addition problem:

3P + 4P + PP + PP = RQ2; where P, Q and R are different digits.

What is the arithmetic mean of all such possible sums?

The correct answer is 202.

Step 1: Express the multi-digit numbers in algebraic form

We are given the addition equation:

$\text{3P}+\text{4P}+\text{PP}+\text{PP}=\text{RQ2}$

Each term can be broken down according to place value as follows:

• $\text{3P}=30+P$
• $\text{4P}=40+P$
• $\text{PP}=10P+P=11P$

Adding the four terms together on the left-hand side:

$\text{Sum}=(30+P)+(40+P)+11P+11P$

$\text{Sum}=70+24P$

Therefore, we have:

$70+24P=\text{RQ2}$

Step 2: Determine possible values for the digit P

The units digit of the sum $\text{RQ2}$ is $2$.

Since $70$ has a units digit of $0$, the units digit of $24P$ (or simply $4\times P$) must end in $2$.

Testing single-digit values of $P$ (where $0\le \mathrm{P<mo>\le </mo><mn>9</mn>}$):

• If $P=3$: $4\times 3=12$ (units digit is 2)
• If $P=8$: $4\times 8=32$ (units digit is 2)

No other digit for $P$ gives a units digit of 2.

Step 3: Test each candidate value of P

Case 1: When $P=3$

$\text{Sum}=70+24\left(3\right)=70+72=142$

Comparing $142$ with $\text{RQ2}$, we get $R=1$ and $Q=4$.
Since $P=3$, $Q=4$, and $R=1$ are all distinct digits, $142$ is a valid sum.

Case 2: When $P=8$

$\text{Sum}=70+24\left(8\right)=70+192=262$

Comparing $262$ with $\text{RQ2}$, we get $R=2$ and $Q=6$.
Since $P=8$, $Q=6$, and $R=2$ are all distinct digits, $262$ is also a valid sum.

Step 4: Calculate the arithmetic mean of all possible sums

The two possible values for the sum are $142$ and $262$.

$\text{Arithmetic Mean}=\frac{142+262}{2}=\frac{404}{2}=202$

Thus, the arithmetic mean of all such possible sums is 202.