for any composite number ... check only the highest power of prime factors divisibility .... i guess that will do the work for example ... 400 = 2^4 * 5^2 so you need to check the divisibility by 25 as well as 16 hope this will meet your need
There are 2 digits in 30. They are 3 and 0.
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.
It would be 2 raised to the power 32: 4,294,967,296.
There are 90 of them.
01
There are seven digits: 8,000,000
for any composite number ... check only the highest power of prime factors divisibility .... i guess that will do the work for example ... 400 = 2^4 * 5^2 so you need to check the divisibility by 25 as well as 16 hope this will meet your need
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.
Seven: 10, 16, 20, 25, 40, 50, 80.
400
It is 2.
10 digits
I'm sure there are more than 2 prime numbers that are 400 digits long.
2 digits are in 30
There are 2 digits in 30. They are 3 and 0.
5 significant digits.