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bonanova

Powerful numbers

Question

The number abc has a decimal representation of ax102 + bx101 + cx100. But let's change things around a bit, and drop the base ten. We could then write abc = a2 + b1 + c0 and ask what values of a b c satisfy the equation. There actually may not be a solution. But if we play with the exponents a bit, we might come up with some numbers that do work. Try these

(The numbers are in order, smallest to largest.)

  1. ab = a2 + b3
  2. cd = c2 + d3
  3. efg = e1 + f2 + g3
  4. hij = h1 + i2 + j3
  5. klm = k1 + l2 + m3
  6. nop = n1 + o2 + p3
  7. qrst = q1 + r2 + s3 + t4
  8. uvwx = u1 + v2 + w3 + x4
  9. yz@$ = y1 + z2 + @3 + $4

There is a shorthand notation. These can also be written ab (2,3), cd (2,3), efg (1,3,5), ... yz@$ (1,2,3,4).

So here's a bonus challenge: abcdefgcc (4, 3, 8, 5, 7, 9, 0, 8, 8)

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2 answers to this question

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Spoiler
  1. 43
  2. 63
  3. 135
  4. 175
  5. 518
  6. 598
  7. 1306
  8. 1676 <-- Note: v and x have the same value...I'm not sure if that's allowed by the OP.
  9. 2427 <-- Note y and @ have the same value...I'm not sure if that's allowed by the OP.

BONUS: Not finding a solution...

      abcdefgcc = a4 + b3 + c8 + d5 + e7 + f9 + g0 + c8 + c8  OR

      abcdefgcc = a4 + b3 + 3c8 + d5 + e7 + f9 + 1

Even with allowing duplicate values for the variables and allowing leading zeros in the number on the left, I'm not finding a solution...

 

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