Albania 2004

[rsoldo]

Let $s(a)$ is sum of divisors of natural number $a$. Prove:
a) If $a=105^{2004}$, then it is $s(a)<\frac{35}{16}a$.
b) Between $18$ consecutive natural numbers exist number $a$ such that $s(a)\geq\frac{13}{6}a$.

P.S.$s(a)=\sigma(a)$ the sum of the divisors is multiplicative arithmetical function and

$$\sigma(a)=(1+p_1+ ... +p_1^{r_1})\cdot (1+p_2+ ... +p_2^{r_2})... = \frac{p_1^{r_1+1}-1}{p_1-1}\cdot \frac{p_2^{r_2+1}-1}{p_2-1}...$$

where is: $a=p_1^{r_1}\cdot p_2^{r_2}....$

[ИСН]

Why, that's trivial. For (a), observe that $\dfrac{\sigma(p^n)}{p^n}<{p\over p-1}$. For (b), among these 18 there will be a number divisible by 18, hence...

[rsoldo]

a) Ma prove for this part:

$105^{2004}=3^{2004}\cdot 5^{2004}\cdot 7^{2004}=(1+3+ ... +3^{2004})(1+5+ ... +5^{2004})(1+7+ ... + 7^{2004})=\frac{3^{2005}-1}{2}\cdot \frac{5^{2005}-1}{4}\cdot \frac{7^{2005}-1}{6}... $

Now we have:

$(3^{2005}-1)\cdot (5^{2005}-1) \cdot (7^{2005}-1)<105\cdot 105^{2004}$
$(3^{2005}-1)\cdot (5^{2005}-1) \cdot (7^{2005}-1)<105^{2005}$
$(3^{2005}-1)\cdot (5^{2005}-1) \cdot (7^{2005}-1)< 3^{2005}\cdot 5^{2005}\cdot 7^{2005}$

and this is obviously true.

[kotenok gav]

rsoldo в сообщении #1461207 писал(а):

Ma prove for this part:

This is absolutely the same as what said ИСН.