, 2021

An integer is divisible by both 12 and 20 exactly when it is divisible by the least common multiple of 12 and 20.

The first few positive multiples of 20 are 20, 40, 60. Since 60 is divisible by 12 and neither 20 nor 40 is divisible by 12, then 60 is the least common multiple of 12 and 20.

Since $60 \cdot 16 = 960$ and $60 \cdot 17 = 1020$, the smallest four-digit multiple of 60 is $60 \cdot 17$.

Since $60 \cdot 166 = 9960$ and $60 \cdot 167 = 10\,020$, the largest four-digit multiple of 60 is $60 \cdot 166$.

This means that there are $166 - 17 + 1 = 150$ four-digit multiples of 60.

Now, we need to remove the multiples of 60 that are also multiples of 16.

Since the least common multiple of 60 and 16 is 240, we need to remove the four-digit multiples of 240.

Since $240 \cdot 4 = 960$ and $240 \cdot 5 = 1200$, the smallest four-digit multiple of 240 is $240 \cdot 5$.

Since $240 \cdot 41 = 9840$ and $240 \cdot 42 = 10\,080$, the largest four-digit multiple of 240 is $240 \cdot 41$.

This means that there are $41 - 5 + 1 = 37$ four-digit multiples of 240.

Finally, this means that the number of four-digit integers that are multiples of 12 and 20 but are not multiples of 16 is $150 - 37 = 113$.