Phat

n=16 (17); n=15 (16); n=0 (49)

So n+1 = 0 (17) = 0 (16) = 0 (272) (I have chosen 272=16×17 because 17 is prime)

272 = 27 (49); 20×272 = 1 (49)

So n = 20 × 272 - 1 = 5440 - 1 = 5439

We have to solve three simultaneous equations

$\boxed{\text{x=16 mod 17; x=15 mod 16; x=0 mod 49}}$

From the first of these equations we have

$x=16+17m\dots (1)$

Substituting this in the second boxed equation gives

$16+17m=15 \text{ mod 16}$

Since we are working modulo 16 we can subtract $16(1+m)$ from the left hand side without affecting the cogruence, so

$m=15 \text{ mod 16}$

Substituting this in (1) gives

$x=16+17\times 15 \text{ mod (17 * 16)}$

$\implies x=271 \text{ mod 272}$

$\implies x=-1 \text{ mod 272}\dots(2)$

Now we use the last of the boxed conditions, which demands that

$x=49y\dots(3)$

Substituting this in (2) gives

$49y=-1 \text{ mod 272}$

The inverse of 49 modulo 272 is 161. I found this using the Euclidean algorithm, and confirmed it using an online modular inverse calculator. So multiplying by 161 gives

$y=-161 \text{ mod 272}$

$\implies y=111 \text{ mod 272}$

This is the smallest possible positive value of y, and so substituting this in (3) gives the smallest possible solution to the boxed equations -

$x=49 \times 111=\boxed{5439}$

First we observe that the positive integer N is congruent to -1 mod 16 and -1 mod 17. Thus we write N as N=16 17 k-1=272k-1 for some positive integer k.

272k-1 = 0 (mod 49) becomes 27k = 1 (mod 49). Thus, we find the inverse of 27 mod 49 which is 20 (you can do this by the extended Euclidean Algorithm or Guess and Check). k=20 is the smallest value that satisfies 27k=1 (mod 49), as it is the inverse, so N=272*20-1=5439.

Let N = 17A-1 = 16B-1 = 49C

Which gives B = (49C+1)/16 = (48C+C+1)/16

This gives minimum C=15 and B=46

Now 17A = 16B = 16(46+49k)

Since 46 is 12mod(17) and 49 is -2mod(17), we take k=6 to get,

17A = 16(46+294) = 5440

Therefore, N = 5440-1 = 5439