Muhammad's skipping list

100=244 (base 6) See that if x is in the yth place in the base 6 representation of n, then the yth place of nth number that is in Eva's list is the xth possible digit (in ascending order of 1,2,4,5,7,8). The 2nd digit that can be used is 2. Therefore, the number in the hundreds place is 2. The 4th digit that can be used is 5. Therefore, the numbers in the tens and ones places are both 5. The 100th number in Eva's list is 255.

From 1-100, there are only 42 which will fit into the list. 1-10 has 6, 11-20 has 6, 21-30 has 6, 41-50 has 6, 71-80 has 6, 81-90 has 6, so we have a total of 42 integers from 1-100 that will fit into the list. For 100-200 however, there is 1 less pair as 101-110 doesn't count into the list. So for 100-200 we have 36 integers that will fit into the list, and now we still need 100-42-30 = 22 more integers that fit in the list. Since 211-220 has another 6, 221-230 another 6, 241-250 another 6, and we still need 22-6-6-6=4 more fitting integers. From there, we count that the 4th integer we need is 251,252,254,255. So 255 is the 42+36+6+6+6+4= 100th integer.

The numbers she wrote don't contain 0,3,6 or 9.So the digits of every number she wrote are from the other six digits.So the total numbers she wrote from 1 to 99 is 6*6=36.The numbers she wrote from 100 to 199 is also 36.Because the first digit of these numbers is 1 which is not divisible by 3.By the same reason the 100th number of the list is 200+(28th number of the list). And the 28th number is 55.So the answer is 255.

number by number I did kkk

We lose the digits 0, 3, 6, and 9. We have six single digit numbers, and 36 two digit number. The 100th number is the 58th three digit number. 58 = 1 6^2 + 3 6^1 + 4*6^0 So that is the 2nd numeral, the 4th numeral and the 4th numeral Which is 255.

Enumerating 1st to 100th term: 1, 2, 4, 5, 7, 8, 11, 12, 14, 15, 17, 18, 21, 22, 24, 25, 27, 28, 41, 42, 44, 45, 47, 48, 51, 52, 54, 55, 57, 58, 71, 72, 74, 75, 77, 78, 81, 82, 84, 85,

We lose the digits 0, 3, 6, and 9. We have six single digit numbers, and 36 two digit number. The 100th number is the 58th three digit number. 58 = 1 6^2 + 3 6^1 + 4*6^0 So that is the 2nd numeral, the 4th numeral and the 4th numeral Which is 255.

Enumerating 1st to 100th term: 1, 2, 4, 5, 7, 8, 11, 12, 14, 15, 17, 18, 21, 22, 24, 25, 27, 28, 41, 42, 44, 45, 47, 48, 51, 52, 54, 55, 57, 58, 71, 72, 74, 75, 77, 78, 81, 82, 84, 85, 87, 88, 111, 112, 114, 115, 117, 118, 121, 122, 124, 125, 127, 128, 141, 142, 144, 145, 147, 148, 151, 152, 154, 155, 157, 158, 171, 172, 174, 175, 177, 178, 181, 182, 184, 185, 187, 188, 211, 212, 214, 215, 217, 218, 221, 222, 224, 225, 227, 228, 241, 242, 244, 245, 247, 248, 251, 252, 254, 255 So the 100th Number of the list is 255.

The digits divisible by 3 are 0, 3, 6, and 9. So every 10 numbers (1 - 10 or 41 - 50) we have 6 allowed numbers.

We also have to consider which groups of ten we can count. For example, 61 - 70 would not work because it has the tens digit 6 which is not allowed (70 does not have the digit 6, however it does have the digit 0 so we can take the whole group of 10 out).

Dividing 100 by 6, we get 16 with a remainder of 4. Counting out 16 approved groups of 10 we arrive to the 17th approved group to count the last four numbers on the list. The 100th number is 255