## Task 4 - 211214

Every natural number can be uniquely represented as a product of prime powers.

In which of the two natural numbers 1981! and 1000! $\cdot$ 981! does the prime number 7 have the larger exponent in this representation?

Hint: If $n$ is a natural number with $n \geq 1$, then $n!$ is defined by $n!=1 \cdot 2 \cdot \ldots \cdot(n-1) \cdot n$.
From the factors 1, 2, ..., 1981 of the number 1981!, since $1981 = 283 \cdot 7$, exactly the 283 numbers

$$
7(=1 \cdot 7) \quad ; \quad 14(=2 \cdot 7) \quad ; \quad \ldots \quad ; \quad 1981(=283 \cdot 7)
$$

are divisible by 7. Therefore, one can divide 1981! by $7^{283}$ by replacing these 283 factors with the numbers 1, 2, ..., 283 (and leaving the other factors in 1981! that are not divisible by 7 unchanged). In this replacement, a product $P$ is formed, in which, because $283 = 40 \cdot 7 + 3$, exactly the 40 numbers

$$
7(=1 \cdot 7) \quad ; \quad 14(=2 \cdot 7) \quad ; \quad \ldots \quad ; \quad 280(=40 \cdot 7)
$$

are still divisible by 7. Therefore, $P$ can be divided by $7^{40}$ by replacing these 40 factors with the numbers 1, 2, ..., 40. Of the factors of the resulting product $Q$, exactly the 5 numbers $7, 14, \ldots, 35$ are divisible by 7. Therefore, $Q$ can be divided by $7^{5}$ again, and in the resulting product, no factor contains the prime number 7 anymore.

Thus, the prime number 7 appears in 1981! with the exponent $283 + 40 + 5 = 328$.

In a similar manner, from

$$
1000 = 142 \cdot 7 + 6 \quad, \quad 142 = 20 \cdot 7 + 2 \quad, \quad 20 = 2 \cdot 7 + 6
$$

it follows that 7 appears in 981! with the exponent $140 + 20 + 2 = 162$. Therefore, 7 appears in 1000! $\cdot$ 981! with the exponent 326 in total. Hence, the prime number 7 has the larger exponent in the representation of the number 1981!.

Adapted from $[5]$

### 9.23.2 Second Round 1981, Class 12