Largest prime is large

We will first obtain the answer, and then justify it.

Because $7$ is a factor of $B-1$ , $B=7k+1$ . Note that because $B$ is prime and greater than $2$ , $k$ must be even. Checking for primality, $B$ can be $29,43,71,113, \ldots$ .

If $B=29,$ then likewise $A = 29k+1$ where $k$ is even. The smallest few possible $A$ are $59,233,349, \ldots$ .

If $B=43,$ then likewise $A = 43k+1$ where $k$ is even. The smallest few possible $A$ are $173,431, \ldots$ .

If $B=71,$ then likewise $A = 71k+1$ where $k$ is even. then the smallest possible $A$ is $569, \ldots$ .

So $A=59,173,233, \ldots$ are the values that are at most 347. If $A=59,$ then the smallest possible $N$ is $709.$ If $A=173,$ then the smallest possible $N$ is $347.$ If $A=233,$ then the smallest possible $N$ is $467.$ So it seems like the smallest value of $N$ is $347.$

To prove that 347 is the smallest possible, suppose that $N$ is smaller. Because $A\leq \frac{N-1}{2},$ we have $A\leq \frac{345-1}{2}=172,$ so $A \leq 171$ . This implies $B\leq 85$ . But we already checked all $B$ and $A$ in this range when we backtracked from $7$ to find our answer.

So the final answer is $347.$

Starting with $B$ , since $B-1$ has its highest prime factor as $7$ , $B$ can be written as $B-1=7n$ or $B=7n+1 \ldots (1)$ . Similarly for $A$ , $A=Bm+1\ldots (2)$ and $N=Ap+1\ldots (3)$ .

Considering equation $(1)$

We know that B is prime, then $n$ should be even.This is the condition, and the value of $n$ should be such that $B$ is prime. Considering the vlaues of $n$ , $n=2, B=15$ not a prime.

$n=4, B=29$ prime,

$n=6, B=43$ prime,

$n=8, B=57$ not a prime,

$n=10,B=71$ prime.

But since they have asked the smallest prime number N , lets consider $B=29$ (smallest value of B) : $A=29m+1$ where m is an even natural number such that A is prime. When

$m=2, A=59$ prime.

$m=4, A=117$ not a prime.

so, considering $A=59$ : $N=59p+1$ where p is an even natural number such that N is prime. when

$p=2,N=119$ , not prime,

$p=4,N=237$ not prime,

$p=6,N=355$ not prime

$p=8,N=473$ not prime

$p=10,N=591$ not prime

$p=12,N=709$ prime.

So, when $B=29$ we get the value of N to be $709$ .

Now, when $B=43$ (next value of B) : $A=43m+1$ , using the similar procedure we find that ,

$m=2, A=87$ not prime.

$m=4, A=173$ which is the smallest value of A being prime.

So, for $A=173$ : $N=173p+1$

The prime values of N is for

$p=2,N= 347$ which is prime.

So, we got value of N which is prime and smaller than $709$ . Now, consider $B=71$ , then value of A will be for

$m=2, A=143$ which is not a prime.

$m=4, A=285$ not a prime.

We can see that all the further taken for $A$ will give a value of $N$ larger than $347$ . This also true for further value of $B$

From this we can come to the conclusion that $347$ is the smallest prime number with the above property or condition. So, the answer is $\boxed {N=347}$

Summarize the information given in 3 equations and several conditions as follows:

1) N-1=A a
2) A-1=B b
3) B-1=7*n

And the conditions are: a,b,n are positive integers; N,A,B are prime.

Starting from equation 3), rearrange this to get:

4) B=7*n+1

Now find the multiples of 7 then add 1 to each:

7*1+1=8

7*2+1=15

7*3+1=22

7*4+1=29

7*5+1=43

7*6+1=43

You do this until you find a prime (since B needs to be prime). As you will notice there are multiple primes, try the smallest first and then the following ones, with the following method:

Using B=29. (7*4+1=29)

Repeat the same process for B=29 using:

5) A=B*b+1

Therefore:

29*1+1=30

29*2+1=59

And 59 is the lowest prime you can make with equation 7), therefore A=59.

For the final step rearrange equation 1):

6) N=A*a+1

Therefore:

59*1+1=60

59*2+1=119

59*3+1=178

59*4+1=237

59*5+1=296

59*6+1=355

. . .

59*12+1=709

This is the lowest value for N you can get with B=29. At first glance it may seem that this has to be the lowest possible value for N, since the lowest possible value for B has been used. However there is the possibility that another value for B will yield a lower value for N.

Therefore, you must repeat the same process for greater values of B as well.

Starting with B=42 (from B=7*6+1) N=347, which is the correct answer.

Note: to be sure you should try every prime number, B, resulting from B=7*n+1. You only have to try values for n up to 35 since after that the resulting values of N would be greater than 1000.

Using Python 2.7.4

def isPrime(n):
    for i in range(2,int(n**0.5)+1):
        if n%i==0:
             return False
    return True

def max_prime_factor(n):
    largest_factor = 1
    i = 2
    while i * i <= n:
        if n % i == 0:
            largest_factor = i
            while n % i == 0:
                n //= i
        i = 3 if i == 2 else i + 2
    if n > 1
        largest_factor = n
    return largest_factor
while True:
    if isPrime(n):
        A=max_prime_factor(n-1)
        B=max_prime_factor(A-1)
        if max_prime_factor(B-1)==7:
                  print n
              break
n+=1

B=7t+1 (t is smallest product of prime numbers that make B a prime no.) thus if t=6, B=43 now A=43m+1 (") thus if m=4,A=173 which is a prime no. now N=173k+1 (") thus if k=2,N=347

Basically, we must first realise that all N, A, and B are prime numbers. To find the smallest possible prime number N, we are inclined to think that A and B are also at the minimum.

B can be expressed in the form of 7p+1. The smallest prime number that fits into this condition would be 43. Hence B is 43.

We then substitute B as a factor of A-1, so A is expressed as 43q+1. By listing, we will find that 347 is the smallest value that is a prime number. Therefore, N has a value of 347.

The first statement tells us that A must be a prime number and the second statement that B is also a prime number. Now the solution must be worked from last statement,i.e, largest prime factor of B-1 is 7, it means that 7 x+1=B (A MULTIPLE OF SEVEN PLUS ONE IS A PRIME NUMBER), where x is a natural number. Since we have to deal with smallest of numbers in this problem, the smallest possibility of B is 43. Now, similarly for the second statement, 43 y+1=A, and from the table of prime numbers, we get to know that the smallest possibility of A is 173. similarly for the first statement, 173*z+1=N, and the smallest possibility in this case is 347, which is N. Hence the answer, N=347

7 has to be a prime factor of B-1. So B can be written as 7b+1. B is the largest prime divisor of A-1, so b has to be an even number and b has to be less than 7, since 7 is the largest prime factor of B-1. For b is 4 B equals the prime number 29. For b = 6 B = 43

Suppose B = 29 is a prime divisor of A-1, then A can be written as 29a + 1 with a less than 29. For a = 2 A = 59, for a = 8 A = 233 and for a = 12 A = 349. Suppose B = 43 then A might be equal to 173 or 431. Since the answer N = An + 1 has to be less or equal 999, larger prime divisors then 499 are irrelevant.

If 59 is the largest prime factor of N-1, then N can be written as 59n + 1. For N a prime number the smallest number of n has to be 12, then N = 709. If 233 is the largest prime factor of N-1, then N = 233n + 1. The smallest number of n then is 2, and N = 467, which is smaller than 709. But if B = 173 then N = 173n + 1 with smallest n equals 2 provides N = 347. Which has to be the smallest prime number since all other options are excluded.