Find all prime numbers p1, p2, p3 … pn such that
i=1∏npi=10i=1∑npi
Given a quadrilateral ABCD with ∠B=∠D=90∘. 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 ∠MHN=∠MCK.
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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?
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Yes
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 p1 and p2 are 2 and 5. We now have
p1p2…pn=10(p1+p2+…+pn)
p3p4…pn=7+p3+p4+…+pn
WLOG p3≤p4≤…≤pn
We will now prove that there is only a maximum of 4 terms in this sequence. Assume that there are k terms for k≥5.
p3p4…pk=p3+p4+…+pk+7
We have p3+p4+…+pk+7≤(k−2)pk+7 and p3p4…pk≤pkk−2. We can prove via induction (or other methods) that (k−2)pk+7<pkk−2 for pk≥3,k≥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)
p3p4=7+p3+p4
(p3−1)(p4−1)=8
We must have p3−1=2 and p4−1=4 so p3=3 and p4=5. Thus, we have solution 2,3,5,5.
Case 2: 3 terms in the sequence
p1p2p3=10(p1+p2+p3)
p3=7+p3
No solutions here.
Case 3: 2 terms in the sequence
p1p2=10(p1+p2)
1=7
No solutions here.
Therefore, only 2,3,5,5 is the solution that satisfies.