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# Olympiad Proof Problem - Day 7

1. Find all prime numbers $$p_{1}$$, $$p_{2}$$, $$p_{3}$$ $$\ldots$$ $$p_{n}$$ such that $\prod_{i=1}^{n} p_{i} = 10 \sum_{i=1}^{n} p_{i}$

2. Given a quadrilateral $$ABCD$$ with $$\angle B=\angle D=90^{\circ}$$. Point $$M$$ is chosen on segment $$AB$$ so that $$AD=AM$$. Rays $$DM$$ and $$CB$$ intersect at point $$N$$. Points $$H$$ and $$K$$ are feet of perpendiculars from points $$D$$ and $$C$$ to lines $$AC$$ and $$AN$$ respectively. Prove that $$\angle MHN=\angle MCK$$.

###### Try more proof problems at Olympiad Proof Problems.

Note by Surya Prakash
2 years, 1 month ago

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Just for clarification, in Q1, can two primes be the same, i.e. is it possible to have 5 and 5 as two of the primes?

- 2 years, 1 month ago

Yes

- 2 years, 1 month ago

Q1) A bit of a sleepy proof (I'm gonna sleep soon). Tell me my errors and I'll rectify them.

Firstly, note that the LHS must be a multiple of 10, so 2 of the primes (we'll call them $$p_1$$ and $$p_2$$ are 2 and 5. We now have

$p_1 p_2 \ldots p_n = 10(p_1+p_2+\ldots+p_n)$

$p_3 p_4 \ldots p_n = 7+p_3+p_4+\ldots+p_n$

WLOG $$p_3 \leq p_4 \leq \ldots \leq p_n$$

We will now prove that there is only a maximum of 4 terms in this sequence. Assume that there are $$k$$ terms for $$k \geq 5$$.

$p_3 p_4 \ldots p_k = p_3 + p_4 + \ldots + p_k+7$

We have $$p_3+p_4+\ldots+p_k +7 \leq (k-2) p_k + 7$$ and $$p_3 p_4 \ldots p_k \leq p_k^{k-2}$$. We can prove via induction (or other methods) that $$(k-2)p_k+7 < p_k^{k-2}$$ for $$p_k \geq 3, k \geq 5$$. Thus, no solutions exist when there are 5 or more terms.

We now have 3 cases:

Case 1: 4 terms in the sequence

$p_1 p_2 p_3 p_4 = 10(p_1+p_2+p_3+p_4)$

$p_3 p_4 = 7 + p_3 + p_4$

$(p_3 - 1)(p_4 - 1) = 8$

We must have $$p_3 - 1 =2$$ and $$p_4-1=4$$ so $$p_3=3$$ and $$p_4 =5$$. Thus, we have solution $$2, 3, 5, 5$$.

Case 2: 3 terms in the sequence

$p_1 p_2 p_3 = 10(p_1+p_2+p_3)$

$p_3 = 7 + p_3$

No solutions here.

Case 3: 2 terms in the sequence

$p_1 p_2 = 10(p_1+p_2)$

$1 = 7$

No solutions here.

Therefore, only $$2, 3, 5, 5$$ is the solution that satisfies.

- 2 years, 1 month ago

Comment deleted Dec 24, 2015

$$\dfrac{3}{5}+\dfrac{7}{5}=2$$ Even I have done such a mistake before...

- 2 years, 1 month ago

Try 2, 3, 5, 5.

- 2 years, 1 month ago

i am having a feeling there is something wrong in this solution....

- 2 years, 1 month ago

The sum can still be an integer. Note that all the fractions on the RHS are not whole numbers since they have prime numerators and denominators. That's where you went wrong.

- 2 years, 1 month ago

oops , i will think and edit in a bit...

- 2 years, 1 month ago