Determine the number of quadruplets of natural numbers $(a, b, c, d)$ satisfying $a b c d=$ 98.

N.B. $(98,1,1,1)$ and $(1,1,98,1)$ are different quadruplets.
Answer: 40. The equation is $a b c d=2 \times 7^{2}$. One of the numbers $a, b, c, d$ is divisible by 2, which gives 4 possible choices. If, for example, it is $a$, then we write $a=2 a^{\prime}$ and the equation becomes $a^{\prime} b c d=7^{2}$. Either two of the numbers $a^{\prime}, b, c, d$ are equal to 7 and the others to 1 (6 possibilities), or one of $a^{\prime}, b, c, d$ is equal to 49 and the others to 1 (4 possibilities). The answer is therefore $4 \times(6+4)=40$.