[superprismatic]

Start at the number 3. Then, at each

step, apply one of eight functions until,

after the final step, you have reached

the target number.

The four functions you may use are:

f₁(x) = x+1

f₂(x) = x-1

f₃(x) = the quotient when x² is divided by 123

f₄(x) = the remainder when x² is divided by123

f₅(x) = x+sod(x)

f₆(x) = x-sod(x)

f₇(x) = x+pod(x)

f₈(x) = x-pod(x)

Note: sod(x) is the sum of digits of x

in its base 10 representation. pod(x)

is the product of digits of x (excluding

leading zeroes) in its base ten

representation. Now, x can be represented

as 123×q+r, where q and r are integers

with r<123. So, f₃(x) would be q

and f₄(x) would be r. All of

the eight functions produce integers.

For example, if I had a target of 148,

I could get there thusly:

Step 1: f₄(3) = 9

Step 2: f₅(9) = 18

Step 3: f₄(18) = 78

Step 4: f₇(78) = 134

Step 5: f₁(134) = 135

Step 6: f₃(135) = 148

Starting at 3, see if you can get to

these targets in as few steps as

possible:

1. 20 (easy)

2. 129 (medium)

3. 271 (hard)

Of course, one could always apply f₁

over and over. This would not, however,

produce the fewest number of steps.

Besides this would be like the

wall-hugging technique to get through

a standard 2 dimensional maze -- it

works, but it is in no way optimal!

[Guest]

f₄(3) = 9

f₇(9) = 18

f₁(18) = 19

f₁(19) = 20

f₄(3) = 9

f₄(6) = 81

f₅(12) = 90

f₄(21) = 105

f₅(72) = 111

f₅(81) = 114

f₅(90) = 120

f₅(99) = 123

f₅(99) = 129

those are my preliminary guesses for now, ill try 271 later.

[superprismatic]

f₄(3) = 9

f₇(9) = 18

f₁(18) = 19

f₁(19) = 20

f₄(3) = 9

f₄(6) = 81

f₅(12) = 90

f₄(21) = 105

f₅(72) = 111

f₅(81) = 114

f₅(90) = 120

f₅(99) = 123

f₅(99) = 129

those are my preliminary guesses for now, ill try 271 later.

Very Nice start!

[Guest]

I have not confirmed if these are optimal as that might take some programming, of which I don't have the software.

1. f₅(3) = f₇(3) = 6

2. f₄(6) = 15

3. f₇(15) = 20

1. f₄(3) = 9

2. f₄(9) = f₅(9) = f₇(9) = 18

3. f₄(18) = 78

4. f₇(78) = 134

5. f₆(134) = 126

6. f₃(126) = 129

1. f₄(3) = 9

2. f₄(9) = f₅(9) = f₇(9) = 18

3. f₄(18) = 78

4. f₇(78) = 134

5. f₁(134) = 135

6. f₇(135) = 150

7. f₃(150) = 182

8. f₃(182) = 269

9. f₁(269) = 270

10. f₁(270) = 271

[Guest]

1. f₄(3) = 9

2. f₂(9) = 8

3. f₄(8) = 64

4. f₇(64) = 88

5. f₇(88) = 152

6. f₃(152) = 187

7. f₃(187) = 284

8. f₆(284) = 270

9. f₁(270) = 271

[superprismatic]

I have not confirmed if these are optimal as that might take some programming, of which I don't have the software.

1. f₅(3) = f₇(3) = 6

2. f₄(6) = 15

3. f₇(15) = 20

1. f₄(3) = 9

2. f₄(9) = f₅(9) = f₇(9) = 18

3. f₄(18) = 78

4. f₇(78) = 134

5. f₆(134) = 126

6. f₃(126) = 129

1. f₄(3) = 9

2. f₄(9) = f₅(9) = f₇(9) = 18

3. f₄(18) = 78

4. f₇(78) = 134

5. f₁(134) = 135

6. f₇(135) = 150

7. f₃(150) = 182

8. f₃(182) = 269

9. f₁(269) = 270

10. f₁(270) = 271

Nice work (including your improvement for 271 in post#5)! You made a mistake in step 2 for the easiest one (20) though.

[Guest]

1. f₁(3) = 4

2. f₄(4) = 16

3. f₂(16) = 15

4. f₇(15) = 20

1. f₅(3) = f₇(3) = 6

2. f₅(6) = f₇(6) = 12

3. f₅(12) = 15

4. f₇(15) = 20

1. f₅(3) = f₇(3) = 6

2. f₅(6) = f₇(6) = 12

3. f₄(12) = 21

4. f₂(21) = 20

1. f₅(3) = f₇(3) = 6

2. f₄(6) = 36

3. f₂(36) = 35

4. f₈(35) = 20

1. f₄(3) = 9

2. f₅(9) = f₇(9) = 18

3. f₁(18) = 19

4. f₁(19) = 20

[plainglazed]

Step 1 - f₄(3) = 9

Step 2 - f₄(9) = 81

Step 3 - f₁(81) = 82

Step 4 - f₇(82) = 98

Step 5 - f₇(98) = 170

Step 6 - f₅(170) = 178

Step 7 - f₃(178) = 257

Step 8 - f₅(257) = 271

[Guest]

This is the optimal solution.

f₁(3) = 4

f₄(4) = 16

f₂(16) = 15

f₇(15) = 20

f₅(3) = 6

f₄(6) = 36

f₂(36) = 35

f₄(35) = 118

f₇(118) = 126

f₃(126) = 129

f₄(3) = 9

f₄(9) = 81

f₁(81) = 82

f₇(82) = 98

f₇(98) = 170

f₅(170) = 178

f₃(178) = 257

f₅(257) = 271

Edited August 3, 2010 by almus