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#### 3.7.59 Programming Tutorial Exercise 2

One way is to move the number to the top of the stack, operate on it, then move it back: `C-x ( M-<TAB> n M-<TAB> M-<TAB> C-x )`.

Another way is to negate the top three stack entries, then negate again the top two stack entries: `C-x ( M-3 n M-2 n C-x )`.

Finally, it turns out that a negative prefix argument causes a command like `n` to operate on the specified stack entry only, which is just what we want: `C-x ( M-- 3 n C-x )`.

Just for kicks, let's also do it algebraically: `C-x ( ' -$$$, $$, $ <RET> C-x )`.