Tag Archives: 2-digit Num3ers

2-digit Numbers: 10*a + b, S=a+b, P=a*b

Let   N   be a 2-digit number: N   =   10*a   +   b Let   S   be the sum of the digits of   N, and   P   the product of the digits of … Continue reading

Prime num3ers under 100 | 10^p – p

2,   3,   5,   7,   11,   13,   17,   19,   23,   29,   31,   37,   41,   43,   47,   53,   59,   61,   67,   71,   … Continue reading

2-digit num3ers – prime factors

Prime numbers less than 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (**)   numbers with exactly 2 distinct prime factors (***) … Continue reading

2-digit num3ers ab, cd; ab – a^2 – b^2 = cd – c^2 – d^2

ab   – a^2   –   b^2   =   k   =   cd   –   c^2   –   d^2 k   >   0 1   =   35   –   3^2   – … Continue reading

Some properties of the two digit numbers

(1) 15,   18,   45   are the only numbers of two digits which by the insertion of zero becomes multiples 105 = 7 * 15        108 = 6 * 18        405 = 9 * 45 If … Continue reading

abc * xyz = ax * by * cz

abc   and   xyz   are 3-digit numbers. ax,   by   and cz   are 2-digit numbers. For example, 567 * 432   =   54 * 63 * 72   =   244944     Here’s a … Continue reading

Posted in Math Beauty, Number Puzzles | | 2 Comments

Sum of six 2-digit numbers and their reverses

13 + 42 + 53 + 57 + 68 + 97   =   330   =   79 + 86 + 75 + 35 + 24 + 31 13^2 + 42^2 + 53^2 + 57^2 + 68^2 + 97^2 … Continue reading

Num3ers N = abcd = (m+a)*(m+b)*(m+c)*(m+d)

With the 2-digit numbers, that is, N = ab; 10*a + b = (m + a) * (m + b) 0 ≤ a, b ≤ 9 If   m = 1,   the solution is:   a = 1, b … Continue reading