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Which two 4-digit numbers multiply to give 4^8 + 6^8 + 9^8?


Genady

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Spoiler

Terms are first three terms of a geometric series a = 48, r = (3/2)8, n = 2

Sum = a (rn+1) -1) / (r -1)

         = 48 ((3/2)24 -1) / ((3/2)8 -1))

         = 48 ((324 - 224) / (38 - 28) * (28 / 224))

         = (324 - 224) / (38 - 28)

         = (316 + 38 28 + 216) (38 - 28) / (38 - 28)

         = 316 + 38 28 + 216                              ......   which puts me back where I started!!!!!

           = (38 + 28)2 - 38 28                           ....... aha!

           = ((38 + 28) + 34 24) ((38 + 28) - 34 24)

           = 8,133 x 5,521

 

Power is failing, so no time to check

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23 minutes ago, sethoflagos said:
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Terms are first three terms of a geometric series a = 48, r = (3/2)8, n = 2

Sum = a (rn+1) -1) / (r -1)

         = 48 ((3/2)24 -1) / ((3/2)8 -1))

         = 48 ((324 - 224) / (38 - 28) * (28 / 224))

         = (324 - 224) / (38 - 28)

         = (316 + 38 28 + 216) (38 - 28) / (38 - 28)

         = 316 + 38 28 + 216                              ......   which puts me back where I started!!!!! Of course.

           = (38 + 28)2 - 38 28                           ....... aha! Yep, complete the square.

           = ((38 + 28) + 34 24) ((38 + 28) - 34 24)

           = 8,133 x 5,521 rather 8,113...  Typo perhaps.

 

+1

 

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+1. However, the "take some time" part can be accomplished in three steps, like this:

Spoiler

216+28*38+316 = 

216+2*28*38+316 - 28*3=

(28+38)- (24*34) =

(28+3+ 24*34)*(28+3- 24*34)

which gives both factors at once.

Is this what you've meant?

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