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Just showing those entries submitted by 'Hudgins': (Click here to show all) A prime number cubed (11) that is a cube in every base (in base 4, 1331=64+48+12+1 in base 10=125=5 cubed, in base 5, 1331=125+75+15+1=216=6 cubed, in base 6 it is 7 cubed, in base 7 it is 8 cubed, etc. This is because it is row 3 of Pascal's triangle (in which every row is a power of 11) and therefore in every base that holds it. It's value in that base showed in base ten is equivilent to the original base number plus 1 cubed. In a formula, 1331(base b)=(b+1)^3. [DeMio and Hudgins]
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