Zaben's Imaginative Math

In his calculations there are 9 digits. They are 0, 1, 2, 4, 5, 6, 7, 8 and 9. $$ So, (100)₉ = 1x9² + 0x9¹ + 0x9⁰ = 81 + 0 + 0 = (81)₁₀ $$ => 81

In order for us to solve this problem we need to eliminate any number that has the number 3. So we are going to eliminate 3,13,23,30 to 39,43,53,63,73,83.93. this give us a total os 19 terms eliminated. So if 1 to 100 had originally 100 terms it will have only 81 since 100 -19=81

Firstly, arrange the 100 numbers in a 10*10 grid, with the first row being 1,2,3,4,5,6,7,8,9,10 and the second row being 11,12,13,14,15,16,17,18,19,20.

Notice that the third row and the third column are to be removed, leaving us a 9*9 grid.

Therefore there are 81 terms.

In his calculation the numbers $3,13,23,30,31,32,33,34,35,36,37,38,39,43,53,63,73,83,93$ don,t exist so a total of $19$ numbers for him don,t exist from $1$ to $100$ so there are $100$ numbers form $1$ to $100$ so total numbers for him $100-19=\boxed{81}$

Using multiplication principle: for a 1-digit number: _ this can be from 0 to 9, excluding 0 and 3, then 8 possibilities. for a 2-digit number: _ _ the first digit can't be a 0 or a 3, and the second digit cannot be a 3, so there are 8 possibilities for the first digit and 9 for the second digit ---> 8x9 = 72 numbers without the 3. for a 3-digit number: _ _ _ this is number 100. ---> 1 number without the 3.

Answer: 8 + 72 + 1 = 81.

Basic Solution: since the thirties have 3, subtract 10 from 100. Each set of ten has a number with 3. So subtract 9, for you already counted the one in the thirties. The answer is 81

19 terms like -3,13,23,30,31,32,33,34,35,36,37,38,39,43,53,63,73,83,93 he wont count. so there will be (100-19)=81 terms

Use all possible combinations for the units' nd tens' place except 3

Both units & tens place can be arranged 9 ways _ _ so 81 ways but if course remove nd add 1 (for 00 - remove & for 100 - add) way which will intern give 81