[Guest]

Let S[x] be the digital root function (also known as the repeated digital sum function), where one adds the digits of positive integer x, then adds the digits of the sum until obtaining a single-digit number. For example, S[975] = 3 because 9 + 7 + 5 = 21 and 2 + 1 = 3.

Given S[a^a] = 2, what is the smallest positive integer that a can be such that a is a perfect power?

Note: a is a perfect power if there exist natural numbers m > 1 and k > 1 such that m^k = a.

[Guest]

Let S[x] be the digital root function (also known as the repeated digital sum function), where one adds the digits of positive integer x, then adds the digits of the sum until obtaining a single-digit number. For example, S[975] = 3 because 9 + 7 + 5 = 21 and 2 + 1 = 3.

Given S[a^a] = 2, what is the smallest positive integer that a can be such that a is a perfect power?

Note: a is a perfect power if there exist natural numbers m > 1 and k > 1 such that m^k = a.

Edited August 12, 2010 by Sumit Modi

[Guest]

Ah I missed these riddles, where did that K Sengupta go to?

128 which is 2⁷

[Guest]

Ah I missed these riddles, where did that K Sengupta go to?

128 which is 2⁷

. 2⁷ is a perfect power, but it is not in the form a^a, but in the form a^b. The value for a must be the same for the base as for the power.

He still visits and submits 'tons' of material to another puzzle site -- www.perplexus.info -- I've also came across some of his submissions on other sites. This puzzle was one that I had submitted on Perplexus a while back. K Sengupta provided a correct answer --- don't peek. Try to solve it without doing so.

Your answer is incorrect

[Guest]

. 2⁷ is a perfect power, but it is not in the form a^a, but in the form a^b. The value for a must be the same for the base as for the power.

He still visits and submits 'tons' of material to another puzzle site -- www.perplexus.info -- I've also came across some of his submissions on other sites. This puzzle was one that I had submitted on Perplexus a while back. K Sengupta provided a correct answer --- don't peek. Try to solve it without doing so.

Your answer is incorrect

so you are asking this:

find 'a' that S[a^a]=2 and a is an integer that is > 1 right?

this makes it probably quite easy to brute force as you just go and check 2², 3³ etc.

seems that 5 will do?

or do you want 'a' to be a perfect power? like in S[perfect power ^{perfect power}] = 2

Edited August 12, 2010 by frester

[Guest]

so you are asking this:

find 'a' that S[a^a]=2 and a is an integer that is > 1 right?

this makes it probably quite easy to brute force as you just go and check 2², 3³ etc.

seems that 5 will do?

or do you want 'a' to be a perfect power? like in S[perfect power ^{perfect power}] = 2

Yes, a is to be a perfect power like in S[(m^k)^(mk)] = 2; where (m^k) = (m^k), with m > 1 and k > 1.

[Guest]

. 2⁷ is a perfect power, but it is not in the form a^a, but in the form a^b. The value for a must be the same for the base as for the power.

He still visits and submits 'tons' of material to another puzzle site -- www.perplexus.info -- I've also came across some of his submissions on other sites. This puzzle was one that I had submitted on Perplexus a while back. K Sengupta provided a correct answer --- don't peek. Try to solve it without doing so.

Your answer is incorrect

Whoops, sorry I was so rushed to get this answered that I forgot that part...

I'm going for brute force on this, two loops for i and j and each time check sodRoot(Math.pow(i,j)*Math.pow(i,j))==2...

18^7=612220032

Dunno if there's a smaller one, I've looked for i and j from 2-19, could be that smaller ones exist, I'll modify the script and try again later...

[Guest]

Whoops, sorry I was so rushed to get this answered that I forgot that part...

I'm going for brute force on this, two loops for i and j and each time check sodRoot(Math.pow(i,j)*Math.pow(i,j))==2...

18^7=612220032

Dunno if there's a smaller one, I've looked for i and j from 2-19, could be that smaller ones exist, I'll modify the script and try again later...

Wouldn't you need Math.pow(i,j) ^ Math.pow(i,j) (i.e. ^ instead of *)? the result is extremely large (a much large integer than I can handle in my programming language).

There has to be a different approach...

[Guest]

^ Ouch, yeah I just noticed, I made a list of all perfect sqares from 2-9999 and most of them are 4 digits, 9801⁹⁸⁰¹ isn't a number my comp can calculate the root sod of any time in this century...

4

8

9

16

27

32

36

49

64

81

125

144

169

196

216

225

243

256

289

324

343

361

484

512

529

576

625

676

729

784

900

961

1000

1089

1156

1331

1369

1444

1521

1600

1681

1849

1936

2025

2048

2187

2197

2209

2304

2401

2500

2601

2704

2744

2809

2916

3025

3125

3136

3249

3364

3375

3481

3721

3844

3969

4096

4225

4356

4489

4624

4761

4900

4913

5041

5184

5329

5476

5625

5776

5832

6561

6724

6859

6889

7056

7225

7396

7569

7744

7776

7921

8000

8100

8192

8281

8464

8649

8836

9025

9409

9604

9801

It's much worse than I thought, I think we might actually have to use our heads to solve this one...

[Guest]

Whoops, sorry I was so rushed to get this answered that I forgot that part...

I'm going for brute force on this, two loops for i and j and each time check sodRoot(Math.pow(i,j)*Math.pow(i,j))==2...

18^7=612220032

Dunno if there's a smaller one, I've looked for i and j from 2-19, could be that smaller ones exist, I'll modify the script and try again later...

S[612220032^612220032] = 9. You can imagine the number 612220032^612220032 you presented must be so huge that there indeed must be, as littlej noted, another approach. There is.

The solution is a much smaller number.

[superprismatic]

The smallest such a is 78125 (= 5⁷);

the next smallest is 161051 (= 11⁵).

S(78125⁷⁸¹²⁵)=2 and S(161051¹⁶¹⁰⁵¹)=2.

Edited August 12, 2010 by superprismatic

[Guest]

The smallest such a is 78125 (= 5⁷);

the next smallest is 161051 (= 11⁵).

S(78125⁷⁸¹²⁵)=2 and S(161051¹⁶¹⁰⁵¹)=2.

How did you get to your answer?

[Guest]

The smallest such a is 78125 (= 5⁷);

the next smallest is 161051 (= 11⁵).

S(78125⁷⁸¹²⁵)=2 and S(161051¹⁶¹⁰⁵¹)=2.

You are correct.

The values of S[m^k] for integers m > 0 and k > 1 are cyclic; and, thus, the values for n for S[n] = 2 can be expressed as the union of the values

(2 + 9x)^{(7 + 6y)} and (5 + 9x)^{(5 + 6y)}, where x and y are integers ≥ 0.

 k: 2   3   4   5   6   7

m

1:  1   1   1   1   1   1  

2:  4   8   7   5   1   2  

3:  9   9   9   9   9   9  

4:  7   1   4   7   1   4  

5:  7   8   4   2   1   5  

6:  9   9   9   9   9   9  

7:  4   1   7   4   1   7  

8:  1   8   1   8   1   8  

9:  9   9   9   9   9   9

Where 2 + 9x = 7 + 6y, we find that where y is an integer, x is not. The equation can be rewritten as x = (5 + 6y)/9; and, as (5 + 6y) modulo 9 results in the cyclic period (5, 2, 8), it has no integer solution. Thus, any values of (2 + 9x)^{(7 + 6n)} can be eliminated as a possible solution for a.

For 5 + 9x = 5 + 6y, 9 is comprised of the prime factors 3 and 3, and 6 is comprised of the prime factors 3 and 2.

The lowest common multiple is 18, therefore a = 5 + 18y.

As S[5 + 18y] = 5, we can use our S[m^k] cyclic table above to note if a can be expressed as a perfect power, that a must either be expressed as (2 + 9x)^{(5 + 6y)} or (5 + 9x)^{(7 + 6y)}. With the expression, 5 + 18y, it is now more manageable to find the solution: (5 + 18•(4320)) = 5⁷ = 78125.

[superprismatic]

How did you get to your answer?

I just wrote a little program to compute S(nⁿ) quickly.

This is easy when you notice that it is equal to S((n(mod 9))ⁿ).

The exponent can be peeled off factor-by-factor. Also, I knew

that a number is a perfect power if and only if the GCD of the

exponents of its prime factorization is not 1. So, I looped over

all perfect powers, a, and computed S(a^a). The exponent

peeling-off process is quite simple: a^b·c(mod 9) is

equal to (a^b(mod 9))^c. I used a few tricks

to keep things from getting too large, but they were just programming

tricks. I hope this makes sense.

[Guest]

I just wrote a little program to compute S(nⁿ) quickly.

This is easy when you notice that it is equal to S((n(mod 9))ⁿ).

The exponent can be peeled off factor-by-factor. Also, I knew

that a number is a perfect power if and only if the GCD of the

exponents of its prime factorization is not 1. So, I looped over

all perfect powers, a, and computed S(a^a). The exponent

peeling-off process is quite simple: a^b·c(mod 9) is

equal to (a^b(mod 9))^c. I used a few tricks

to keep things from getting too large, but they were just programming

tricks. I hope this makes sense.

Wow, yes it does make sense. Thanks.

[Guest]

This should not be taken to imply that the digital root is necessarily a multiplicative function. I use an excel spreadsheet for minor puzzle things like digital root as well, but I don't use the MOD function.