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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
Posted in Math Beauty, Number Puzzles
Tagged 2-digit Num3ers, Digit Sum, digital product
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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
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
2-digit & 3-digit Num3ers are k times the sum of the squares of their digits
For example, 10 = 10 * (1^2 + 0^2) (k = 10) that is, if N = ab is a 2-digit number, N = 10*a + b = k * (a^2 + b^2) k > … Continue reading
Interesting Properties for Num3ers between 90 & 100
Here’s a method for multiplying two numbers between 90 and 100: Take, for example, 94 & 97 Add the two numbers, subtract 100 from the sum: 94 + 97 – 100 = 91 Then write … Continue reading
