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.