Solution

Let’s consider the sequence . Note that But Consequently, our inequality holds if and only if n ≤ 2000. This means that the sequence up to the 2001th term increases and then decreases.

Answer

For n = 2001.

Hint

Solution

a$)$ We note immediately that the sequence $a_k$ is strictly increasing. Indeed, the assumption $a_k$ = $a_{k + 1}$ = n immediately leads to a contradiction: $a_n$ = 3k = 3 $(k + 1)$. In addition, $a_1$ $>$ 1 $($otherwise $a_{a_1}$ = $a_1$ = 1 ≠ 3$)$. It follows that $a_k$ $>$ k for all k. On the other hand, $a_1$ $<$ $a_{a_1}$ = 3. Therefore, $a_1$ = 2, $a_2$ = 3, $a_3$ = 6, $a_6$ = 9, $a_9$ = 18, $a_{18}$ = 27, $a_{27}$ = 54, $a_{54}$ = 81, $a_{81}$ = 162, $a_{162}$ = 243. And since 162 - 81 = 243 - 162, then $a_k$ = 81 + k for all k from 81 to 162. In particular, $a_{100}$ = 181.$\\$ b$)$ $a_{162}$ = 243, $a_{243}$ = 486, $a_{486}$ = 729. Since 729-486 = 486 - 243, then $a_k$ = 243 + k for all k from 243 to 486. In particular, $a_{418}$ = 661, which means that $a_{661}$ = 1254, $a_{1254}$ = 1983, $a_{1983}$ = 3762.

1. Similarly, we can obtain the general formula: $\\$
2. In classes 7 and 8, only part a$)$ was offered $($12 points$)$, in classes 9 and 10 only part b$)$ was offered $($8 points$)$.

Answer

a$)$ 181; b$)$ 3762.

Hint

Solution

Answer: For k = 65. We denote $B_k$ = $19^k$/k!, $C_k$ = $66^k$/k!. Then $A_k$ = $B_k$ + $C_k$, $B_{k + 1}$/$B_k$ = 19/$(k + 1)$, $C_{k + 1}$/$C_k$ = 66/$(k + 1)$. Consequently, for k $\leq$ 19, both sequences do not decrease, and for k $\geq$ 65 both sequences do not increase, that is, the maximum value is attained for some k $\in$ [19, 65]. Note that for k $\in$ [19, 64], the inequalities $C_{k + 1}$/$C_k$ $\geq$ 66/65 and $B_k$/$C_k$ = $(19/66)^k$ $<$ $(1/3)^19$ $<$ 1/65 are fulfilled. Consequently, $A_{k + 1} - A_k = C_{k + 1} - C_k + B_{k + 1} - B_k \geq 1/65 C_k - B_k $>$ 0$, that is, for k $\in$ [19, 65] the sequence $A_k$ increases. Hence, the value of $A_k$ is maximal when k = 65.

Hint

Solution

Nothing to translate.

Hint

Solution

Let us prove that for any numbers $a_1, …, a_n$ satisfying the condition of the problem, one can find a number $a_{n + 1}$ such that $S_{n+1} = a_1^2+ …+ a_{n+1}^2$
is divisible by $s_{n + 1} = a_1 + … + a_n + a_{n + 1}$. It follows from the equality that $S_{n + 1}$ is divisible by $s_{n + 1}$ if $S_n+ s_n^2$ is divisible by $s_{n + 1}$, since $a_{n + 1} + s_n = s_{n + 1}$. Thus, it suffices to take – in this case $S_n+ s_n^2 = s_{n+1}$. In this case

Answer

See the solution above.

Hint

Solution

a$)$ Example 1. Take the number 2 and 1024 numbers equal to ½. Then $a_n$ = $2^n$ + $1024 \times 2^{-n}$ = 32 $(2^{n-5} + 2^{5-n})$. The sum of two positive reciprocal inverses is smaller, the closer they are to each other. Therefore, the constructed sequence decreases until n = 5, and then increases.

Example 2. Take $a_n$ = $1.02^n + 0.5^n$. Note that $a_{n + 1} > a_n ⇔ 0.02 \times 1.02^n > 0.5^{n + 1} ⇔ 2.04^n > 25 ⇔ n ≥ 5$. That is, the sequence decreases until n = 5, and then increases.

b$)$ Suppose this happens. The first method. Then $a_n$ $<$ $a_5$ for any n. Among the initial numbers was the number x $>$ 1, otherwise the sequence would never increase. Notice that
$a_n$ $>$ $x^n$ for any n. But the sequence $x^n$ is not bounded. So we have a contradiction.

The second method. Then $a_4$ $<$ $a_5$ and $a_6$ $<$ $a_5$, that is, $a_4$ + $a_6$ $<$ $2a_5$. Hence, $x^4$ + $x^6$ $<$ $2x^5$ for at least one of the original numbers x. But this contradicts the Cauchy inequality.

Answer

a$)$ It could; b$)$ it could not.

Hint

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Solution

a) Denote by p the probability that this target is struck, that is, it is an affected target. By symmetry, the average number of affected targets is np.

Note that the probability that this boy will hit this target is 1/n. Therefore, the probability that this boy will not hit this target is 1 – 1/n. Therefore $($due to the independence of shots$)$ the probability that no boy will hit this target is $(1 – 1/n)^n$. Hence, the probability that at least one boy will hit this target is $1 – (1 – 1/n)^n$.

Thus, the average number of targets hit is $n(1 – (1 – 1 / n)^n)$.

b) We show that the average number of targets hit is greater than n/2. For this, it suffices to show that $(1 – 1 / n)^n $<$1$. Note that the sequence is increasing. Indeed, the sequence $b_n$ of inverse numbers decreases $($see the solution of problem number 61394 c$)$.

In addition, consequently , $($see problem number 60873$)$.

Answer

a) $n(1 – (1 – 1/n)^n)$; b) No, it cannot.

Hint

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Solution

Nothing to translate

Hint

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