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For a prime number $p$ and integers $0\le k\le n$, the P-adic valuation of the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ equals the number of arithmetic carries in the base-$p$ addition of $k$ and $n-k$.^[ref] Indeed, subtract the two denominator factorial valuations in the Legendre formula:Each arithmetic carry accounts for one unit of this quotient by subadditivity of the base-p digit sum. In particular, the binomial coefficient is divisible by $p$ exactly when at least one carry occurs.

For $n=50,k=17,p=3$, the factorial counts giveThe matching base-p expansions are $17=(0122{)}_3$ and $33=(1020{)}_3$. Their units add to two; the next column gives $2+2=4=1+3$, sending one carry into the nines column. That column then gives $1+0+1=2$, and no further carry occurs. The result is $(1212{)}_3=50$. The base-p digit sums check the same result: $(5+3-6)/(3-1)=1$. Consequently $\left(\genfrac{}{}{0ex}{}{50}{17}\right)$ is divisible by three but not by nine.

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

Let $\mu =C_p(\lambda )$, put $m=|\mu |$, and let the first-level $p$-quotient partitions have sizes $n_0,\dots ,n_{p-1}$. ThenRepeated subadditivity of the base-p digit sum givesFor 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 givesApply this to every first-level quotient partition. Their core towers concatenate to levels $r\ge 1$ of $T{C}_p(\lambda )$, so

Part a applied to $\lambda$ and to its $p$-core $\mu$, whose higher core-tower levels are empty, now givesTherefore the Character-degree valuation does not increase on taking the p-core: