# 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2021 Canadian Senior Mathematics Contest Solutions

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2021 Canadian Senior Mathematics Contest Solutions

Since these divisors can be used in any order, there are 17 such sequences, as in (c).
Case $3: 27$ and 36
The divisor 36 can only be used once, otherwise too many factors of 2 would be removed. Since $36 \times 27^{15}=2^2 \times 3^{47}$ and $36 \times 27^{16}=2^2 \times 3^{50}$ and $36 \times 27^{17}=2^2 \times 3^{53}$, then the divisor of 27 must be used exactly 16 times.
Therefore, Leistra sequences can be formed by using exactly 16 divisors equal to 27 and 1 divisor equal to 36 .
Since these divisors can be used in any order, there are 17 such sequences, as in (c).
Case 4: 27 and two $18 \mathrm{~s}$
Note that 18 cannot be used more than two times and that 18 cannot be combined with 12 or 36 , otherwise the remaining quotient would be either odd or not an integer.
Since $18^2 \times 27^{14}=2^2 \times 3^{46}$ and $18^2 \times 27^{15}=2^2 \times 3^{49}$ and $18^2 \times 27^{16}=2^2 \times 3^{52}$, then the divisor of 27 must be used exactly 15 times.
Therefore, Leistra sequences can be formed by using exactly 15 divisors equal to 27 and 2 divisors equal to 18 .
There are $\left(\begin{array}{c}17 \ 2\end{array}\right)=\frac{17 \times 16}{2}=136$ ways of choosing 2 of the 17 positions in the sequence of divisors for the 18 s to be placed. The remaining spots are filled with $27 \mathrm{~s}$.
Therefore, there are 136 such sequences.
Case 5: 27 and one 18
Since $18 \times 27^{15}=2 \times 3^{47}$ and $18 \times 27^{16}=2 \times 3^{50}$ and $18 \times 27^{17}=2 \times 3^{53}$, then the divisor of 27 must be used exactly 16 times, leaving a final term of $2^2=4$.
As in Case 2 and Case 3 , there are 17 such sequences.
Having considered all possibilities, there are $17+17+136+17=187$ Leistra sequences with $a_1=2^3 \times 3^{50}$.

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2021 Canadian Senior Mathematics Contest Solutions

(a) Using $f(x+y)=f(x) g(y)+g(x) f(y)$ with $x=y=0$, we obtain $f(0)=2 f(0) g(0)$.
From this, we get $f(0)(2 g(0)-1)=0$.
Thus, $f(0)=0$ or $g(0)=\frac{1}{2}$.
Using $g(x+y)=g(x) g(y)-f(x) f(y)$ with $x=y=0$, we obtain $g(0)=(g(0))^2-(f(0))^2$.
If $g(0)=\frac{1}{2}$, we obtain $\frac{1}{2}=\frac{1}{4}-(f(0))^2$ which gives $(f(0))^2=-\frac{1}{4}$.
Since $f(0)$ is real, then $(f(0))^2 \geq 0$ and so $(f(0))^2 \neq-\frac{1}{4}$.
Thus, $g(0) \neq \frac{1}{2}$, which means that $f(0)=0$.
Since $f(a) \neq 0$ for some real number $a$, setting $x=a$ and $y=0$ gives
$$f(a+0)=f(a) g(0)+g(a) f(0)$$
Since $f(0)=0$, we obtain $f(a)=f(a) g(0)$.
Since $f(a) \neq 0$, we can divide by $f(a)$ to obtain $g(0)=1$.
Therefore, $f(0)=0$ and $g(0)=1$.
(We note that the functions $f(x)=\sin x$ and $g(x)=\cos x$ satisfy the given conditions, so there does exist at least one Payneful pair of functions.)

# 滑铁卢数学竞赛代考

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2021 Canadian Senior Mathematics Contest Solutions

$18 \times 27^{15}=2 \times 3^{47}$ 和 $18 \times 27^{16}=2 \times 3^{50}$ 和 $18 \times 27^{17}=2 \times 3^{53}$ ，那么 27 的除数必须正好使用 16 次，留下最后一项 $2^2=4$.

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2021 Canadian Senior Mathematics Contest Solutions

(a) 使用 $f(x+y)=f(x) g(y)+g(x) f(y)$ 和 $x=y=0$ ，我们获得 $f(0)=2 f(0) g(0)$. 由此，我们得到 $f(0)(2 g(0)-1)=0$.

$$f(a+0)=f(a) g(0)+g(a) f(0)$$

(我们注意到函数 $f(x)=\sin x$ 和 $g(x)=\cos x$ 满足给定条件，因此确实存在至少一对 Payneful 函 数。)

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