<p>if 6 times j is 1 more than the square of k is an integer , what is the smallest possible value of j ?</p>
<p>We have 6j + 1 = k^2. I’m assuming j >= 1 (otherwise j = 0 works). The LHS is congruent to 1 (mod 2) and 1 (mod 3), so k cannot be a multiple of 2 or 3. Also k is odd. Trying k = 5, we have j = 4.</p>
<p>it’s 6j + 1 = k^2 not 6j = k^2 +1</p>
<p>I thought it would be 6j=k^2 + 1 if it were 6j=k^2, then 6j would be equal to one more than k^2, so to make them equal, you must add one to the k^2</p>
<p>satman 1111 the answer is 1/6 ?? i still didn’t get it ??</p>
<p>Sorry I might have misread the question (it has horrible grammar). If it is read as “6j is 1 more than the square of an integer k” then we would have 6j = k^2 + 1, and the minimum value of j is 1/6.</p>