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 A^2 = B^3 + C^3
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 smallest integer whose first n multiples all contain a 3
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 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: February 2012
Powers & Permutation of Digits
Prove that the list of the numbers with the property described below is finite. I claim that you cannot find a kdigit number (k ≥ 12) with this property. N.B. There are less than 50,000 which have your property (in … Continue reading
Powers & Sum Digits – Num3ers from 2 to 99 [Part 1]
Part One: From the numbers 2 to 99 Next : Numbers from 100 to 200 I’ve checked all the numbers from 2 to 99 raised to the power k, with k = 1, 2, 3, …, 20 2^6 = 64 … Continue reading
4digit Num3ers & Reversals
8712 is an integral multiple of its reversal, 2178, as 8712 = 4 * 2178 Question: Find another 4digit number which is a nontrivial integral multiple of its reversal.
Powers & Sum Digits [Part 2]
Find others Paul found:
Numbers p; p = (p+1)^k + (p+1)^k – (p+2)^k
I’ll be looking for numbers p such that p = (p + 1)^k + (p + 1)^k – (p + 2)^k If k = 2, then p = (p + 1)^2 + (p + 1)^2 – (p + 2)^2 p^2 … Continue reading
Numbers of the form ab * (a+b) = a^k + b^k
ab * (a+b) = a^k + b^k where ab is a 2digit number (= 10*a + b) If k = 2, then 10 a^2 + 11 ab + b^2 = a^2 + b^2 Solution for the variable b: b = … Continue reading
Concatenation, Squares & Sums [Part 1]
(1) Of the form, AB = A  B = A^k + B^k 1,233 = 12  33 = 12^2 + 33^2 1000*a + 100*b + 33 = (10*a + b)^2 + 33^2 The only acceptable solutions are: a = … Continue reading
Ascending & Descending Powers
The set of these numbers is infinite. But the set of numbers that can be expressed in ascending and descending powers seems rare. The search is on! I’m particularly interested in finding prime numbers that can be expressed in sums … Continue reading
Powers are the same as the digits & the reverse
(1) Numbers where the powers are the same as the digits of the numbers 746 = 1^7 + 2^4 + 3^6 986 = 1^9 + 2^8 + 3^6 (2) Numbers where the basenumbers are the digits of the numbers For … Continue reading
Digits (Part 1)  Numbers of the form abc*(ab + c) = (ab)^3 + c^3
To find numbers of the form… abc * (ab + c) = (ab)^3 + c^3 where ab = 10, 11, …, 99 and x = 0, 1, 2, …, 9 abc is a 3digit number, and ab, a … Continue reading