[Guest]

Suppose we call the sum of the digits of the sum of the digits of a whole number its "sodsod. "

For example, the sodsod of 48 is 3 (4+8 = 12; 1+2 = 3).

Now, consider all the four-digit whole numbers:

1) What's the maximum sodsod?

a) Prove it's the maximum (without listing all of them).

2) How many four-digit whole numbers have this maximum sodsod?

a) How many (four-digit numbers with the maximum sodsod) have a 1 in the thousands column?

b) How many have a 2 in the thousands column?

c) a 3?

d a 4?

e) a 5?

f) a 6?

g) a 7?

h) an 8?

i) a 9?

j) Any patterns in a-i above?

[Guest]

The first part is easy,

11. why? you know that the first sum has to be less than 36 (4*9) Therefore, the largest number you can make is 29 which adds to 11.

[k-man]

The largest 4-digit number is 9999, so the largest sod of all 4-digit numbers is 36. All sods of 4-digit numbers are between 4 and 36.

The largest sod of numbers in the range 4-36 is for the number 29 and is 11. So the answer to the first part is

1) the maximum sodsod of all 4-digit numbers is 11.

I don't have the answer to the second question yet, but the answers for 2a through 2e are as follows:

2a) no numbers start with 1 as the smallest number that has the sod of 29 is 2999.

2b) 1 (2999)

2c) 3 (3899,3989,3998)

2d) 6 (4799,4889,4898,4979,4988,4997)

2e) 9 (5699,5789,5798,5879,5888,5897,5978,5987,5996)

Don't have time right now to continue...

[Guest]

First note that by definition the sodsod must be single digit number. Take a four digit number, N, and sodsod(N)= K then the sodsod(N+1)=K+1 unless K=9, which requires on additional step since 9+1=10 making sodsod = 1 for K=9. This will give a repetive pattern of numbers 1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9 etc for all sodsods of consecutive 4 digit numbers. Max sodsod = 9, 1000 sodsod =1,2,3,4,5,6,7,8 or 9. no sodsod =0

MattB and k-man need to read definition again. Note: sodsod(2999)= 2

[Guest]

Okay, maybe I am the one that misinterpreted the intent. I assumed the intention was to continue to 1 digit but that is not exactly what was said, but rather take only two steps.

Greatest sodsod is 11, occuring 120 times.

10 occurs 780 times, 3,4,5,6,7,8 & 9 each occur 1000 times,

2 occurs 880 times and

1 occurs only 220 times.

Proof for 11 highest: Doing the first step will derive highest number of 36 (ie all 9's).

For the second step to be a two digit number will require first step results to be between 19 and 29. Since 29 obviously has the highest integer parts, the addition of those must be the maximum, ie 11.

[Guest]

First note that by definition the sodsod must be single digit number. Take a four digit number, N, and sodsod(N)= K then the sodsod(N+1)=K+1 unless K=9, which requires on additional step since 9+1=10 making sodsod = 1 for K=9. This will give a repetive pattern of numbers 1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9 etc for all sodsods of consecutive 4 digit numbers. Max sodsod = 9, 1000 sodsod =1,2,3,4,5,6,7,8 or 9. no sodsod =0

MattB and k-man need to read definition again. Note: sodsod(2999)= 2

thoughtfulfellow, SODSOD(2999) = 11. xamdam defines SODSOD as the digital sum of the digital sum. What you incorrectly interpreted as SODSOD is called the digital root.

(1) The maximum sum of the digits of four digits is 36 (9+9+9+9), which is a two-digit number. The maximum sum of the digits of two digits <= 36 is 11 (2+9, thus where SOD(n) = 29). Permitting leading zeroes, the first digit of a two-digit number is 0, 1, 2 or 3. The maximum digit of the second digit is the 9, yet the maximum digit where the first digit is 3 is 6, which in the two-digit total is only 9. 2+9 is larger, thus 11 is the maximum.

(2a) There are 120 numbers with a 'SODSOD' of 11.

(2b) 1.

(2c) 3.

(2d) 6.

(2e) 10.

(2f) 15.

(2g) 21.

(2h) 28.

(2i) 36.

(2j) Yes. the number of times the decimal digits 1 thru 9 appear in the thousands column for

four digit (non-leading zero) numbers corresponds to the triangular numbers: a(n) = C(n+1,2) = n(n+1)/2 (offset by 1).

Edited July 5, 2011 by Dej Mar

[Guest]

Maximum sodsod would vary.

for single and 2 digit , it would be 9 ( as 9 is is the maximum for single digit and 99 =18=9 for 2 digits)

but when 3 digits, it would be 10( as 991=19=10)

4 digit, it would be 11( 9992=29=11)

5 digit , it would be 12( 99993=39=12)

6/7/8/9/10/11 digit,it would be 13/14/15/16/17/18

and from 12 digit it starts again with max sodsod of 10 and so on

So maximum sodsod comes to be 18.

[Guest]

The first part is easy,

11. why? you know that the first sum has to be less than 36 (4*9) Therefore, the largest number you can make is 29 which adds to 11.

Yep, I got that too.

[Guest]

The largest 4-digit number is 9999, so the largest sod of all 4-digit numbers is 36. All sods of 4-digit numbers are between 4 and 36.

The largest sod of numbers in the range 4-36 is for the number 29 and is 11. So the answer to the first part is

1) the maximum sodsod of all 4-digit numbers is 11.

I don't have the answer to the second question yet, but the answers for 2a through 2e are as follows:

2a) no numbers start with 1 as the smallest number that has the sod of 29 is 2999.

2b) 1 (2999)

2c) 3 (3899,3989,3998)

2d) 6 (4799,4889,4898,4979,4988,4997)

2e) 9 (5699,5789,5798,5879,5888,5897,5978,5987,5996)

Don't have time right now to continue...

5969 for 2e

[Guest]

thoughtfulfellow, SODSOD(2999) = 11. xamdam defines SODSOD as the digital sum of the digital sum. What you incorrectly interpreted as SODSOD is called the digital root.

(1) The maximum sum of the digits of four digits is 36 (9+9+9+9), which is a two-digit number. The maximum sum of the digits of two digits <= 36 is 11 (2+9, thus where SOD(n) = 29). Permitting leading zeroes, the first digit of a two-digit number is 0, 1, 2 or 3. The maximum digit of the second digit is the 9, yet the maximum digit where the first digit is 3 is 6, which in the two-digit total is only 9. 2+9 is larger, thus 11 is the maximum.

(2a) There are 120 numbers with a 'SODSOD' of 11.

(2b) 1.

(2c) 3.

(2d) 6.

(2e) 10.

(2f) 15.

(2g) 21.

(2h) 28.

(2i) 36.

(2j) Yes. the number of times the decimal digits 1 thru 9 appear in the thousands column for

four digit (non-leading zero) numbers corresponds to the triangular numbers: a(n) = C(n+1,2) = n(n+1)/2 (offset by 1).

I got this too. I liked those triangular numbers. http://en.wikipedia....iangular_number

[Guest]

The first part is easy,

11. why? you know that the first sum has to be less than 36 (4*9) Therefore, the largest number you can make is 29 which adds to 11.

[Guest]

welldone!!!

[Guest]

hi i'm new here and my name is guppy i found this site a few days ago and i enjoyed the puzzles and the forums o i thought i'ld join nice to meet u

now for the second part u can find how many there are if u answer the rest which my answers were

a)u can't have 1 since the smallest is 2999

b)1 (2999)

c)3 (3899 3989 3998)

d)6 (4799 4979 4997 4889 4898 4988)

e)10 (5699 5969 5996 5888 5789 5879 5798 5987 5978 5897)

f)15 (6599 6986 6887 6779 .... etc)

i think i spotted the pattern so i'm just going to go with it i'm seeing is that u just keep adding by plus one for example 1+2=3 and 3+3=6 and 6+4=10 and 10+5=15 so

g)15+6=21

h)21+7=28

i)28+8=36

j) as to the patter i explained my reasoning and i checked it out with with a few more number this is all that makes sense

as to the first ques back up on how many there are we just add the total of our findings (1+3+6+10....=120) so 120 the answer

i didn't look at the answers so i'm just hoping this makes sense

[Guest]

hi i'm new here and my name is guppy i found this site a few days ago and i enjoyed the puzzles and the forums o i thought i'ld join nice to meet u

now for the second part u can find how many there are if u answer the rest which my answers were

a)u can't have 1 since the smallest is 2999

b)1 (2999)

c)3 (3899 3989 3998)

d)6 (4799 4979 4997 4889 4898 4988)

e)10 (5699 5969 5996 5888 5789 5879 5798 5987 5978 5897)

f)15 (6599 6986 6887 6779 .... etc)

i think i spotted the pattern so i'm just going to go with it i'm seeing is that u just keep adding by plus one for example 1+2=3 and 3+3=6 and 6+4=10 and 10+5=15 so

g)15+6=21

h)21+7=28

i)28+8=36

j) as to the patter i explained my reasoning and i checked it out with with a few more number this is all that makes sense

as to the first ques back up on how many there are we just add the total of our findings (1+3+6+10....=120) so 120 the answer

i didn't look at the answers so i'm just hoping this makes sense

Hey, not bad guppy! Welcome to the Den.

[Guest]

thx xamdam glad to be here:D