Numbers such that concatenation of prime factors is a power

 
 
For example,

10 \; = \; 2 \; \times \: \; 5
25 \; = \; 5^2

14 \; = \; 2 \; \times \: \; 7
27 \; = \; 3^3

20 \; = \; 2 \; \times \: \; 2 \; \times \: \; 5
225 \; = \; 15^2

86 \; = \; 2 \; \times \: \; 43
243 \; = \; 3^5

129 \; = \; 3 \; \times \: \; 43
343 \; = \; 7^3

145 \; = \; 5 \; \times \: \; 29
529 \; = \; 23^2

178 \; = \; 2 \; \times \: \; 89
289 \; = \; 17^2

183 \; = \; 3 \; \times \: \; 61
361 \; = \; 19^2

203 \; = \; 7 \; \times \: \; 29
729 \; = \; 3^6

394 \; = \; 2 \; \times \: \; 197
2197 \; = \; 13^3

403 \; = \; 13 \; \times \: \; 31
1331 \; = \; 11^3

802 \; = \; 2 \; \times \: \; 401
2401 \; = \; 7^4

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged , . Bookmark the permalink.

1 Response to Numbers such that concatenation of prime factors is a power

  1. David @InfinitelyManic says:

    Repost …

    (10,[2,5],25,2)
    (14,[2,7],27,3)
    (20,[2,5],25,2)
    (28,[2,7],27,3)
    (40,[2,5],25,2)
    (50,[2,5],25,2)
    (56,[2,7],27,3)
    (80,[2,5],25,2)
    (86,[2,43],243,5)
    (98,[2,7],27,3)
    (100,[2,5],25,2)
    (112,[2,7],27,3)
    (129,[3,43],343,3)
    (145,[5,29],529,2)
    (160,[2,5],25,2)
    (172,[2,43],243,5)
    (178,[2,89],289,2)
    (183,[3,61],361,2)
    (196,[2,7],27,3)
    (200,[2,5],25,2)
    (203,[7,29],729,2)
    (203,[7,29],729,3)
    (203,[7,29],729,6)
    (224,[2,7],27,3)
    (250,[2,5],25,2)
    (320,[2,5],25,2)
    (344,[2,43],243,5)
    (356,[2,89],289,2)
    (387,[3,43],343,3)
    (392,[2,7],27,3)
    (394,[2,197],2197,3)
    (400,[2,5],25,2)
    (403,[13,31],1331,3)
    (448,[2,7],27,3)
    (500,[2,5],25,2)
    (549,[3,61],361,2)
    (640,[2,5],25,2)
    (686,[2,7],27,3)
    (688,[2,43],243,5)
    (712,[2,89],289,2)
    (725,[5,29],529,2)
    (784,[2,7],27,3)
    (788,[2,197],2197,3)
    (800,[2,5],25,2)
    (802,[2,401],2401,2)
    (802,[2,401],2401,4)
    (896,[2,7],27,3)
    (1000,[2,5],25,2)

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