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 k-digit number (k ≥ 12) with this property. N.B. There are less than 50,000 which have your property (in … Continue reading

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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

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4-digit Num3ers & Reversals

8712 is an integral multiple of its reversal, 2178, as 8712 = 4 * 2178 Question: Find another 4-digit number which is a non-trivial integral multiple of its reversal.

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Powers & Sum Digits [Part 2]

    Find others     Paul found:          

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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

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Numbers of the form ab * (a+b) = a^k + b^k

ab * (a+b) = a^k + b^k where ab is a 2-digit 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

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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

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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

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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

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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 3-digit number, and ab, a … Continue reading

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