a=\frac{62k+1}{k-1} or a = [62k+1] / [k-1] a, k are natural numbers I must find all a
Nov 1, 2011 #1 Patronas 3 0 [tex]a=\frac{62k+1}{k-1} [/tex] or a = [62k+1] / [k-1] a, k are natural numbers I must find all a Last edited by a moderator: Nov 6, 2011
General Math News on Phys.org Hermit 'scribblings' of eccentric French math genius unveiled Sperm swimming is caused by the same patterns that are believed to dictate zebra stripes Many Wordle users cheat to win, says mathematics expert
Nov 1, 2011 #2 dacruick 1,042 1 what do you mean you must find all a. Through what method must you find all a?
Nov 1, 2011 #3 mathman Science Advisor 8,140 571 For starters k must be even. If k is odd, k - 1 is even and 62k + 1 is odd.
Nov 1, 2011 #4 Patronas 3 0 I know that k must be even. But then, I calculate for k=2, k=4, k=6 ... And I have results a=63, 65, 69, 71, 83, 125 but i need good method
I know that k must be even. But then, I calculate for k=2, k=4, k=6 ... And I have results a=63, 65, 69, 71, 83, 125 but i need good method