Olympiad Proof Problem - Day 7

  1. Find all prime numbers p1p_{1}, p2p_{2}, p3p_{3} \ldots pnp_{n} such that i=1npi=10i=1npi\prod_{i=1}^{n} p_{i} = 10 \sum_{i=1}^{n} p_{i}

  2. Given a quadrilateral ABCDABCD with B=D=90\angle B=\angle D=90^{\circ}. Point MM is chosen on segment ABAB so that AD=AMAD=AM. Rays DMDM and CBCB intersect at point NN. Points HH and KK are feet of perpendiculars from points DD and CC to lines ACAC and ANAN respectively. Prove that MHN=MCK\angle MHN=\angle MCK.


Try more proof problems at Olympiad Proof Problems.

Note by Surya Prakash
3 years, 10 months 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?

Sharky Kesa - 3 years, 10 months ago

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Yes

Surya Prakash - 3 years, 10 months ago

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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 p1p_1 and p2p_2 are 2 and 5. We now have

p1p2pn=10(p1+p2++pn)p_1 p_2 \ldots p_n = 10(p_1+p_2+\ldots+p_n)

p3p4pn=7+p3+p4++pnp_3 p_4 \ldots p_n = 7+p_3+p_4+\ldots+p_n

WLOG p3p4pnp_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 kk terms for k5k \geq 5.

p3p4pk=p3+p4++pk+7p_3 p_4 \ldots p_k = p_3 + p_4 + \ldots + p_k+7

We have p3+p4++pk+7(k2)pk+7p_3+p_4+\ldots+p_k +7 \leq (k-2) p_k + 7 and p3p4pkpkk2p_3 p_4 \ldots p_k \leq p_k^{k-2}. We can prove via induction (or other methods) that (k2)pk+7<pkk2(k-2)p_k+7 < p_k^{k-2} for pk3,k5p_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

p1p2p3p4=10(p1+p2+p3+p4)p_1 p_2 p_3 p_4 = 10(p_1+p_2+p_3+p_4)

p3p4=7+p3+p4p_3 p_4 = 7 + p_3 + p_4

(p31)(p41)=8(p_3 - 1)(p_4 - 1) = 8

We must have p31=2p_3 - 1 =2 and p41=4p_4-1=4 so p3=3p_3=3 and p4=5p_4 =5. Thus, we have solution 2,3,5,52, 3, 5, 5.

Case 2: 3 terms in the sequence

p1p2p3=10(p1+p2+p3)p_1 p_2 p_3 = 10(p_1+p_2+p_3)

p3=7+p3p_3 = 7 + p_3

No solutions here.

Case 3: 2 terms in the sequence

p1p2=10(p1+p2)p_1 p_2 = 10(p_1+p_2)

1=71 = 7

No solutions here.

Therefore, only 2,3,5,52, 3, 5, 5 is the solution that satisfies.

Sharky Kesa - 3 years, 9 months ago

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