# Chapter 4: Practice Makes Perfect

A huge part of learning recursion is practice! Please try the following problems; there are solutions in Chapter 5 to check your work. Keep in mind that these can be pretty difficult!

```
####
# 1) we'll start off with something familiar. without looking at sum digits,
# implement the following function so that md(number) returns the PRODUCT of the
# digits of number.
#
# md(4023) -> 4 * 0 * 2 * 3 = 0
# md(423) -> 4 * 2 * 3 = 24
####
def md(number):
if __________:
return _____
_____________________
_____________________
return __________________________
```

```
####
# 2) Exponents are basically repeated multiplication! For example,
# 2^3 (2 to the power of 3) = 2 * 2 * 2 = 8
# Basically, it's three 2's multiplied together.
# We can write a function for this recursively!
# 2 is the base, and 3 is the exponent in this case.
# So, for base ^ exponent, write the following function that returns that value!
#
# HINT: remember when we did factorial, we thought about how we have to
# make the problem smaller, and tried to relate the smaller problem
# to our original problem. That may help here!
#
# rec_power(2, 3) = 2 * 2 * 2 = 8
# rec_power(4, 2) = 4 * 4 = 16
#
####
def rec_power(base, exponent):
if ________________:
return ________
exponent = ___________________
return _______________________
```

```
####
# 3) Implement the following function so that count8(number)
# counts the number of times the number "8" appears in number.
#
# I've given you parts of the base cases; if the number left is 8, we've
# found an 8, so what should we return?
# If the number left isn't 8, but can't be made any smaller, what should
# we return then?
#
# count8(3283) -> 1
# count8(32883) -> 2
# count8(8388) -> 3
####
def count8(number):
if number == 8:
return _____
elif number < 10:
return _____
rightmost_digit = _________________
rest_of_number = __________________
if rightmost_digit == 8:
return _______________________
else:
return _______________________
```

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