[math-fun] Doubly true equalities
Hello Math-fun, those equalities are doubly true : 2+3=5 5+1!=6 because they have respectively 5 and 6 symbols, altogether. I'm looking for a general formula, always true for any integer "a". Here is one for any odd "a" (with a > 1): a + 0 + 0 + ... [k times the string "+0"]... + 0 = a with k=(a-1)/2 - L(a) and L(a) being the "length of a" (quantity of digits of integer a). Examples: 97 + 0 + 0 + 0 ... [46 times "+0"] ... + 0 = 97 99 + 0 + 0 + 0 ... [47 times "+0"] ... + 0 = 99 101 + 0 + 0 + 0 ... [47 times "+0"] ... + 0 = 101 103 + 0 + 0 + 0 ... [48 times "+0"] ... + 0 = 103 Here is one for any even "a" (with a > 6): a-1+1! + 0 + 0 ... [k times "+0"] ... + 0 = a with k=(a-6)/2 - L(a) Examples: 98-1+1!+0+0 ... [44 times "+0"] ... +0=98 100-1+1!+0+0 ... [44 times "+0"] ... +0=100 102-1+1!+0+0 ... [45 times "+0"] ... +0=102 The general formula could be of any shape, of course -- no need to play with "+0" strings or factorials. Best, É.
Hi Eric, It seems to me that since you used 1! to represent 1 (apparently to get one more symbol), you can do things with parentheses, 0!, 1^2 (either with a caret or a superscript), 0^2, 1^[2^(3^4)], etc., to get an arbitrarily large number of symbols. Kerry On Nov 22, 2007 10:30 AM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Hello Math-fun,
those equalities are doubly true : 2+3=5 5+1!=6 because they have respectively 5 and 6 symbols, altogether.
-- lkmitch@gmail.com www.fractalus.com/kerry
I have a key-ring calculator where the number symbols are intact but the operation symbols have worn off. So to rediscover which operation key is which, I enter 3, operation key, 2, Enter. If 1 appears, it was -; if 5, +; if 6, x; etc. Nothing brilliant about that, but this post reminded me of it. I'm sure this notion could be generalized into a nontrivial logical puzzle. Steve Gray Kerry Mitchell wrote:
Hi Eric,
It seems to me that since you used 1! to represent 1 (apparently to get one more symbol), you can do things with parentheses, 0!, 1^2 (either with a caret or a superscript), 0^2, 1^[2^(3^4)], etc., to get an arbitrarily large number of symbols.
Kerry
On Nov 22, 2007 10:30 AM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Hello Math-fun,
those equalities are doubly true : 2+3=5 5+1!=6 because they have respectively 5 and 6 symbols, altogether.
I have an old scientific calculator that I changed the battery on, but the keys fell out in the process & I had to put them back in more-or-less random order. So I have the same "puzzle" ! This old Casio has, in addition to the usual +-*/ functions, the trig & statistical functions. At 09:51 AM 11/22/2007, Steve Gray wrote:
I have a key-ring calculator where the number symbols are intact but the operation symbols have worn off. So to rediscover which operation key is which, I enter 3, operation key, 2, Enter. If 1 appears, it was -; if 5, +; if 6, x; etc. Nothing brilliant about that, but this post reminded me of it. I'm sure this notion could be generalized into a nontrivial logical puzzle.
Steve Gray
Kerry Mitchell wrote:
Hi Eric,
It seems to me that since you used 1! to represent 1 (apparently to get one more symbol), you can do things with parentheses, 0!, 1^2 (either with a caret or a superscript), 0^2, 1^[2^(3^4)], etc., to get an arbitrarily large number of symbols.
Kerry
On Nov 22, 2007 10:30 AM, Eric Angelini <Eric.Angelini@kntv.be> wrote:
Hello Math-fun,
those equalities are doubly true : 2+3=5 5+1!=6 because they have respectively 5 and 6 symbols, altogether.
participants (4)
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Eric Angelini -
Henry Baker -
Kerry Mitchell -
Steve Gray