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 A^2 = B^3 + C^3
 Set {4,b,c,d,e} such that the product of any two of them increased by 1 is a square
 smallest integer whose first n multiples all contain a 3
 Set {3,b,c,d,e} such that the product of any two of them increased by 1 is a square
 Set {2,b,c,d,e} such that the product of any two of them increased by 1 is a square
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Monthly Archives: July 2012
The Last Two Digits of the Square of some Num3ers
Notice the pattern 13^2 = 169 14^2 = 196 the last two digits of the square of 13 are interchanged, the result is the … Continue reading
Prime Num3er 41 and its 5digit multiples
41 is a prime number. Let abcde be a multiple of 41, its digitrotation numbers: bcdea, cdeab, deabc and eabcd are also multiples of 41 For example, 41 * … Continue reading
Fun puzzle: 4digit Num3er abcd
Find a 4digit Num3er abcd such that (1) a + b = c + d the sum of the first two digits is equal to that of the last two digits. (2) a + … Continue reading
a + b + … + n = a * n
Sum of consecutive integers is equal to to the product of the first and the last terms: a, b, c, …, n are consecutive integers. a + b + … + n = a * n For example, 3 … Continue reading
Challenging question: Sum of Twin Primes
A twin prime is a prime number that differs from another prime number by two, we write (p, p+2) The first few twin prime pairs are: (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), … Continue reading
Prime Num3ers Average
The prime numbers under 1000 are: 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, 101, 103, 107, 109, 113, 127, 131, 137, 139, … Continue reading
a^4 + b^4
Did you know that a^4 + b^4 = (a^2 + ab 2 + b^2) (a^2 – ab 2 + b^2) http://en.wikipedia.org/wiki/Factorization a^4 + 4 b^4 = (a^2 + 2 b^2)^2 – (2ab)^2
Sum of 2digit Numbers is Sum of Squares/Cubes of their digits
(1) Squares Let ab and cd be 2digit number, then let’s look for 2digit number such that (10*a + b) + (10*c + d) = a^2 + b^2 + c^2 + d^2 which gives me a^2 + 10*a … Continue reading
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Prime Numbers formed by 2 (or more) consecutive Fibonacci numbers in reverse order
Fibonacci numbers: F(n) = F(n1) + F(n2) with F(0) = 0 and F(1) = 1 http://en.wikipedia.org/wiki/Fibonacci_number 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, … Continue reading
Powers that start with 3 & 4 & 5 identical digit
The squares: a(n) = n^2 for n = 0 … 10,000 The cubes: a(n) = n^3 for n = 0 … 10,000 4th powers: a(n) = n^4 for n = 0 … 1,000 5th powers: a(n) = … Continue reading