One is the digit that appears the most as each number appears an equal amount of times e.g teens, twenties, etc. However as the final number is 9001, the extra one causes one to be the most occurring. Zero is the least occurring as zero is always after another number, never before. So A=1 B=0 , therefore 1+0 =1

First, we notice 1-9 are used the same time N in the sequence 1-9999 and 0 is M as M<N.

We can easily find N which is 4000. For M we have:

1. from 1 to 9 it's 0
2. from 10 to 99 it's 9
3. from 100 to 999 it's 9 $\times$ 20 (for each 1XX, 2XX, ..., 9XX) + 9 (from point 2)) = 189
4. from 1000 to 9999 it's 9 $\times$ 300 (for each 1XXX, 2XXX, ..., 9XXX) + 189 (from point 3)) = 2889

So, finally M=2889.

Now we have to remove the digits from the sequence 9002-9999. We have:

• for 1: N - 300 (for 9XXX) + 1 (for 9001) = 4000 - 300 + 1 = 3701
• for each 2,3,..., 8: N - 300 (for 9XXX) = 4000 - 300 = 3700
• for 9: N - 1300 (for 9XXX) + 2 (for 9000 and 9001) = 4000 - 1300 + 2 = 2702
• for 0: M - 300 (for 9XXX) + 3 (for 9000) + 2 (for 9001) = 2889 - 300 + 3 + 2 = 2594

Finally, 1 is used $\boxed{3701}$ times (max value) and 0 is used $\boxed{2594}$ times (min value) which makes $\boxed{a=1}$ and $\boxed{b=0}$ so a + b = $\boxed{1}$ .

For programmers the following C code directly shows a and b :

#include <cstdlib>
#include <iostream>

using namespace std;

int main()
{
int t[10] = {0};
int i, j;

for(i=1;i<=9001;i++){
  j=i;
  while(j){
    t[j%10]++;
    j/=10;         
  }      
}

for(i=0;i<10;i++)
  printf("%d=[%d] ", i, t[i]);
printf("\n");

return 0;
}

The output is:

0=[2594] 1=[3701] 2=[3700] 3=[3700] 4=[3700] 5=[3700] 6=[3700] 7=[3700] 8=[3700] 9=[2702]

Think of a number as this : _ _ _ _ that is, four places to be filled up with digits.

Now certainly, to give each digit its fair share of presence, we should start with 0000 and go up to 9999. Doing so, we would do justice to every digit.

But bored as he was, Fred din't have the energy to enter all those useless zeros as prefixes ( for example, he wrote 24 instead of 0024). So admittedly, he did a grave injustice to zeros, making them the most scarce digit on the board. On the other hand, starting with 1 and ending with 1 in the unit's place gave 1 an edge, making it the most abundant one.

So the most and least abundant digits add up to 1.

Even without doing the detailed calculations, it is easy to observe that if we consider all the numbers from 1 to 8999, there would be an equal number of 1's, 2's, 3's, ..., 7's and 8's, fewer 9's and much lesser 0's (0 NOT in the leading digits). Even after considering the two left out numbers i.e. 9000 and 9001, 0 would still appear the least of the times and 1 would appear the most of the times. a = 1 and b = 0 ==> a + b = 1 + 0 = 1.

The calculations are given below: Total Number of 0's = {0 + 9 + 2 * (9 * 10) + 3 * (8 * 10 + 10)} + 5 = (0 + 9 + 180 + 2400) + 5 = 2589 + 5 = 2594

Total Number of 1's = [1 + (10 + 9) + {10 * 10 + 2 * (9 * 10)} + {10 * 10 * 10 + 3 * (8 * 10 * 10)}] + 1 = (1 + 19 + 280 + 3400) + 1 = 3700 + 1 = 3701

Total Number of 2's, 3's, ..., 8's = [1 + (10 + 9) + {10 * 10 + 2 * (9 * 10)} + {10 * 10 * 10 + 3 * (8 * 10 * 10)}] = 1 + 19 + 280 + 3400 = 3700

Total Number of 9's = [1 + (10 + 9) + {10 * 10 + 2 * (9 * 10)} + {3 * (8 * 10 * 10)}] + 2 = (1 + 19 + 280 + 2400) + 2 = 2700 + 2 = 2702

From 1 to 99 every number is same no of times....... except one O in the begining

From 100 to 999 also it is same.....

From 1000 to 8999 is also same

So we have 9000 and 9001

9 = 2 times

0 = 3+2-1(at the begining) = 4 times

1 = 1 times

So a = 0 .......... b = 1

Sum = a+b = 1

$1$ appears the most because it is up to $9001$ -- meaning that the "1" appears as a units digit one more time than the rest of the digits 1-9. $0$ appears the least because the starting digit cannot be $0$ .

0 is the least occurring (obviously), so we just find the most occurring. All the numbers have the same amount of digits (except 9) till 9000. 9001 contains an extra 1, so 1 is the most occurring.

1+0=1

Much simpler answer. Let's look at what happens at each digit.

1st: 1~0 loop + 1 "So it ends with extra 1"

2nd: 1~0 loop

3rd: 1~0 loop

4th: 1~0 loop + 1~9 "My goodness 0 is missing!"

a=1, b=0

a+b = the answer~~!

as Fred is listing between 1 to 9001 ,in wrinting all these the most appearing digit will be 1(you can think it by just supposing integers range from 1 to 20 .in this range 1 appear for 11 times and 0 for 2 times only .by doing this practice u will observe 1 is the most occuring disgit and 0 is the least so sum of 1+0 is 1

1-8 digit would be same in numbers till 9000. Then for 9001, 1 becomes highest in number 0 will always be less than other 9 digits in numbers as it never appears as the digit in most significant place.

1. 2 digits - [1-9][0-9]
2. 3 digits - [1-9][0-9][0-9]
3. 4 digits - [1-9][0-9][0-9][0-9]

For numbers from 1 - 9000, the first digit can be 1-9, second, third and fourth digits are also 0-9. This means 1-9 appears equal times. Since it ends at 9001, there is an extra 1.

fred listed integers from 1 to 9001.. the most used digit is zero... because in thousands 3 or 2 zeros are used. eg: in thosand, 1000 three zeros are used and in 1001,1002,1003,.... 2 zeros are used! therefore the dgit he wrote the most 'a'= 0 , and the digit he wrote the least is 1,, because he wrote integers upto 9001... not 10000,10,001... therefore 'b'=1,,, now a+b=0+1=1

1 to 9001 = 9001. 9001 = 9000+1 = a+b= 9000+(1). 1 is the correct answer.

digit 1 appears the most and digit 0 appears the least number of times from 1 to 9001. for example: from 1 to 101; digit 1 appears 23 times and digit 0 appears 12 times and the rest digit appears less than 23 times and more than 12 times. So, a is 1 and b is 0 hence a+b=1+0=1

I didn't even see the problem when I solved this. I thought it was asking for the digit that appears the most in the picture and I saw 1. LOL!

1

Most frequent digit ->0(a) [ By educated guess or intuition] Least frequent digit-> 1(b) [ same as above] a+b= 1

0 occurs least as 1-100 are not written as 001,002 etc. till 9000 digits 1-8 occurs same. therefore by 9001 , digit 1 occurs once again.

Because 1 appears more than any digit, for the sequence starts and ends with 1 as a digit, it must be a. Since no number will start with 0, it is the least used and is b.

1+0=1.

1+0

1

x 10 20 30 . . . . . . . 90 100
1 11 21 31 . . . . . . . 91
2 12 22 32 . . . . . . . 92
3 . . . . . . . . . . . . . . . .
: :
: :
9 19 29 39 . . . . . . . .99

>no of 2's are - 20
>no of 3's, 4's ....9's are 20 .
>no of 1's = 20 + 1(from 100) = 21
>no of 0's = 9 + 2 (from 100) = 11
>this is the case from 1 to 100 , similarly the numbers are in the same ratios from 1 to 9000 plus we are getting an additional " 1 " from 9001.
>so a = 1 and b = 0

1 to 99 =20 100 to 199 =120 200 to 299 =20 300 to 399 =20 400 to 499 =20 500 to 599 =20 600 to 699 =20 ...................( my hand is cramping :( 900 to 999 =20 1000 = 1 total=301

so, from 1 to 1000: digit '1' appears 301 times.

Therefore 1 to 9001 digit '1' appears more times than other digits

From 1 to 9, 11 to 99, 111 to 999, 1111 to 9999 all the digit appear at the same quantity. But 0 and 1 are still appear in 10, 100, 1000. And 9999 decrease to 9001, all the digit from 2 to 9 decrease at the same quantity except 0 and 1 decrease fewer. So the answes is 1+0=1

when writing a set of 100 integers starting with one, you write every digit exactly the same number of times as the other except zero. so, obviously no matter how much integers you write, you will write zero the least number of times. hence we have our value b=0.

now "a". consider the set of hundred integers i mentioned above. every digit will be written 20 times in it. expanding this set to a 1000 integers we write the numbers 1-100 with 1,2,3.....and 9 in front of them. like, 100 ones in 100-199 100 twos in 200-299 . . . . 100 nines in 900-999

a little consideration with this result, we understand that we write the digits 1-9 the same number of times no matter how many numbers we write, provided we finish at a thousand or million and so on.

here in our problem fred has written till 9001, one more than a thousand. so the only digit written the most is "1". it has been written one more time than the digits 2-8.

so a=1.

now we get a+b=1.

NOTE: in this problem, 9 is not written the same number of times as the digits 2-8 because, the numbers 9002-9999 are not written, anyway zero is written fewer number of times than nine. here we need not worry about it.

b=0, and since he wrote till 9001 a =1