Counting and Summing

[dtweah]

Start with the integers from 1 to 10^2006. Replace each of them by the sum of its digits to get a string of 10^2006 numbers. Keep doing this until you get 10^2006 single digit numbers. Let m be the number of 1^s and n be the number of 2's. Then m-n equals?

A. 0

B. 1

C. 2

D. 3

E. None of the above

[m&m]

I think the answer is surprisingly E (I've never had a "none of the above" on a GMAT question)


to get sum of 1, it's essentially all the factors of 10..i ie 1, 10, 100, 1000, 10000, etc.
there are 2006 terms

to get sum of 2, we could have form where 2 is leading ie, 2, 20, 200, 2000, etc. as above so 2006 terms
OR
we could have 11,110,101,1001,1010,1100, etc. which has even MORE than 2006 terms

so m-n must be Very negative since n >> m hence E

What's OA?

[vitaly]

"Keep doing this until you get 10^2006 single digit numbers"

So for example 92 will be first 9+2=11 and then 1+1=2? Then solution from m&m doesn't work

[dtweah]

I think the answer is surprisingly E (I've never had a "none of the above" on a GMAT question)


to get sum of 1, it's essentially all the factors of 10..i ie 1, 10, 100, 1000, 10000, etc.
there are 2006 terms

to get sum of 2, we could have form where 2 is leading ie, 2, 20, 200, 2000, etc. as above so 2006 terms
OR
we could have 11,110,101,1001,1010,1100, etc. which has even MORE than 2006 terms

so m-n must be Very negative since n >> m hence E

What's OA?

OA follows after few more responses

[Uri]

i think the answer is 1

1.....2......3......4.....5.......6......7......8........9
10...11....12....13....14....15....16.....17......18
19...20....21....22....23....24....25.....26......27
observe that all the numbers in the same column produce the same integer when their constituent digits are summed up. if we continue this way, we will find that ultimately 100 comes just below 1. so, 10^any number will also come just below 1. until that final number, the total number of 1's and 2's will be same. so when that final number (10^2006 in this question) is taken into account, we will have one more 1 than we will have 2.

[vitaly]

Uri, that's great!

[dtweah]

i think the answer is 1

1.....2......3......4.....5.......6......7......8........9
10...11....12....13....14....15....16.....17......18
19...20....21....22....23....24....25.....26......27
observe that all the numbers in the same column produce the same integer when their constituent digits are summed up. if we continue this way, we will find that ultimately 100 comes just below 1. so, 10^any number will also come just below 1. until that final number, the total number of 1's and 2's will be same. so when that final number (10^2006 in this question) is taken into account, we will have one more 1 than we will have 2.

Good job Uri. OA is 1. After every 9 terms, total 1s and 2 the same. There are 10^2006 numbers. (10^2006)/9 has remainder of 1. To see it,
(9+1)^2006. There is a 9 in every expansion except the last which is 1.