This can be solved by looking at each set of digit lengths. From 1 through 9, it's obvious there are 9 digits. From 10 through 99, there are 2 digits for each of the 90 numbers, so that makes 90*2=180 digits. Next, from 100 to 400, there are 401 numbers with 3 digits each, making another 401*3=1,203 digits. So the final answer is 9+180+1,203 = 1,392 digits.
1 in 1 and 3 in 200.
1 to 9 have 1 digit and 10 to 99 have 2 digits so (9 x 1) + (90 x 2) = 189
1,2,3,4 1+2+3+4=10 4 times 3 times 2 times 1 =24 24 counting numbers
10
2,401
The sum of the number of digits in all the numbers between 31 and 400 inclusive is 1041.
121
Four: 516,561,615 & 651.
There are 2892 digits.
192 digits
I believe here are 51 such numbers.
There are 2700 digits.
11522
There are 400. Assuming the number must be at least 10,000, then: In a 5 digit palindrome, the first and last digits must be the same, and the second and fourth digits must be the same; and: For the first and last digit there is a choice of 4 digits {2, 4, 6, 8}; For each of these there is a choice of 10 digits {0, 1, ..., 9} for the second and fourth digits; For each of the above choices these is a choice of 10 digits {0, 1, ..., 9} for the third digit; Making 4 x 10 x 10 = 400 possible even 5 digit palindromes.
Infinity is not a number so we would not know how many digit would be in infinity it just something that you can tell endless that is why infinity was invented
400 ÷ 1/4 = 400 × 4/1 = 1600
101