Add the base-$p$ expansions of $x$ and $y$ column by column. Before carrying, the sum of all displayed digits is $d_p(x)+d_p(y)$. Each carry removes $p$ units from one column and adds one unit to the next, decreasing the total digit sum by $p-1$. After all carries the digits are those of $x+y$, so the <subadditivity of the base-p digit sum> gives
$$
\boxed{d_p(x+y)\leq d_p(x)+d_p(y)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 4 / b / Solution

Let $\mu=C_p(\lambda)$, put $m=|\mu|$, and let the first-level $p$-quotient partitions have sizes $n_0,\ldots,n_{p-1}$. Then
$$
n=m+p\sum_jn_j.
$$
Repeated <subadditivity of the base-p digit sum> gives
$$
d_p(n)-d_p(m)
\leq d_p\left(\sum_jn_j\right)
\leq\sum_jd_p(n_j).
$$
For any partition $\nu$ of size $s$, iterating the core-quotient relation $s=|C_p(\nu)|+p|Q_p(\nu)|$ and the same digit-sum inequality gives
$$
d_p(s)\leq\sum_{r\geq0}|TC_p(\nu)_r|.
$$
Apply this to every first-level quotient partition. Their core towers concatenate to levels $r\geq1$ of $TC_p(\lambda)$, so
$$
d_p(n)-d_p(m)
\leq\sum_{r\geq1}|TC_p(\lambda)_r|.
$$

Part a applied to $\lambda$ and to its $p$-core $\mu$, whose higher core-tower levels are empty, now gives
$$
\begin{aligned}
(p-1)\left(v_p(\chi^\lambda(1))-v_p(\chi^\mu(1))\right)
&=\sum_{r\geq1}|TC_p(\lambda)_r|-d_p(n)+d_p(m)\\
&\geq0.
\end{aligned}
$$
Therefore the <Character-degree valuation does not increase on taking the p-core>:
$$
\boxed{v_p(\chi^\lambda(1))\geq v_p(\chi^{C_p(\lambda)}(1))}.
$$

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 4 / c / Solution

For nonnegative integers $x$ and $y$,
$$
d_p(x+y)\leq d_p(x)+d_p(y).
$$
In column addition, every carry replaces $p$ units in one digit by one unit in the next digit and therefore decreases the digit sum by $p-1$.


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 Incoming links: Subadditivity of the base-p digit sum