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# 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)

## 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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Nice work. Bonus solution is the one I had in mind.

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