Prime, prime, prime, prime, prime, prime
There exists a unique six-term sequence of prime numbers, p 1 , p 2 , . . . , p 6 such that p i + 1 = 2 p i + 1 and p 1 < 1 0 0 . Find p 1 .
The answer is 89.
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6 solutions
primes from one to one hundred:
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97
if the first prime first digit is 1,3,7,then
3-7-5-out
7-5-out
1-3-7-5-out
The primes remaining are:
19,29,59,79,89
19,59 remain 1 when divided by three( 1 × 2 +1=3),so they can't be the solution
The next number after 29 is 59,after 59 it will be 119,a multiple of 7
Therefore the answer is 89
testemos a congruência no módulo dos 3 primeiros primos: mod 2: Pi♡1(mod 2)
mod 3: se Pi♡1(mod 3), então Pi+1♡0(mod 3) absurdo! ; se Pi♡2(mod 3), então Pi+1♡2(mod 3) ok!
mod 5: se Pi♡1(mod 5), então Pi+1♡3(mod 5) (i) ; se Pi♡2(mod 5), então Pi+1♡0(mod 5) (ii) ; se Pi♡3(mod 5), então Pi+1♡2(mod 5) (iii) ; se Pi♡4(mod 5), então Pi+1♡4(mod 5) (iv) ; o (ii) é absurdo, implicando que o (iii) também é um absurdo, que implica q o (i) também é um absurdo, logo o único não absurdo é o (iv) ; logo, pelo teorema do resto chinês, temos que: Pi♡29(mod30), logo, nesse caso temos q Pi será ou 29 ou 59 ou 89 ; testemos a congruência módulo 7: 29♡1(mod 7), então 2x29+1♡3(mod 7), ou seja, 59♡3(mod 7), então, 59x2+1♡0(mod 7), ou seja, 119♡0(mod 7), logo Pi não pode ser nem 29, nem 59, logo Pi=89
sorry, but english please
write in portuguese , actually there are others solutions in english , u don't need to write in english , muito pelo contrário haha
write in english
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This solution has the same approach as the other.
2 doesn't work. 2, 5, 11, 23, 47, 95. Start with a prime ending with 1, 3, 5, 7 and it doesn't work as the last digit will be 5 before the 6th number in the sequence. Only 9 will work. The last digit remains 9 for all numbers in sequence. 19, 29, 59, 79, 89 are options. Hit and Trial leads to 89.
We understand from Pi = 2Pi + 1 that the numbers in the sequence will be 1 added to double the previous number
As we know, numbers with units digits as 7 could not be the first term of the sequence as those numbers will give the next term to be a number which has its units digit as 5. Numbers with units digit as 5 cannot be prime. Therefore, numbers with units digit as 7 are ruled out from being the first term. we are given that Pi < 100 We start checking with P1 = 2 and we get 95 as a term in our way, we check with P1 = 3 and we get 15 in our way. Fail We then start with prime numbers below 100, ignoring 97, we reach 89 We start checking with 89 as first term and we get rest of the terms as prime numbers There P1 = 89
Since P1 is prime, P1 can only end in 1,3,7,9 and also 2 and 5 (if P1 is 2 or 5).
If our P1 ends in 7, when doubled and added 1, we will get a number ending in 5 but greater than 5, so it cannot be prime. Ergo, P1 cannot end in 7.
If our P1 ends in 3, P2 will end in 7, and P3 will end in 5. Using the same reasoning as above, P1 cannot end in 3.
If our P1 ends in 1, P2 will end in 3, P3 will end in 7, and P4 will end in 5; P1 cannot end in 1.
If our P1 is 5, P2 ends in 1, P3 ends in 3, P4 ends in 7, P5 ends in 5; P1 cannot be 5.
If our P1 is 2, P2 ends in 5, P3 ends in 1, P4 ends in 3, P5 ends in 7, P6 ends in 5; P1 cannot be 2.
That leaves us with a P1 which ends with 9. Prime numbers less than 100 ending in 9 are 19, 29, 59, 79, and 89.
Trying them one by one,
If P1 is 19, P2 is 39, NOT PRIME
If P1 is 29, P2 is 59, P3 is 119. NOT PRIME (div. by 7)
If P1 is 59, P3 is 119. NOT PRIME (div. by 7)
If P1 is 79, P2 is 159. NOT PRIME (div. by 3)
So P1 is obviously 89 since that's the only thing we have left (we don't need to calculate for the other terms anymore XD)
Anyway, the other terms are as follows, P1 = 89, P2 = 179, P3 = 359, P4 = 719, P5 = 1439, P6 = 2879, We can verify that all of them are prime.