There are 100 even numbers between 1 and 200 (inclusive).
As the numbers are between 200 and 300 then the first and last digits must be 2 ; the numbers must be 2?2. The possible palindromic numbers in this range are 202, 212, 222, 232, 242, 252, 262, 272, 282 and 292. The difference between any two of these numbers is a number between 10 and 90 which is a multiple of 10.
There are 43 natural numbers between 200 and 500 that are divisible by seven.
I think the answer is: 66 numbers between 1 and 200 are divisible by 3. :) That's what I think the answer is.
Required numbers from 200 and 300 are 207,217,227,237,247,257,267,270,271,272,273,274,275,276,278,279,287,297 these are the numbers therefore, the answer is 18
10
202, 212, 222, 232, 242, 252, 262, 272, 282, and 292 are the palindromic numbers between 200 and 300. NONE of them are prime, since they must all end in 2 they are even and so are divisible by 2, so they are not prime.
Infinitely many numbers are below 200.
There are 100 even numbers between 1 and 200 (inclusive).
As the numbers are between 200 and 300 then the first and last digits must be 2 ; the numbers must be 2?2. The possible palindromic numbers in this range are 202, 212, 222, 232, 242, 252, 262, 272, 282 and 292. The difference between any two of these numbers is a number between 10 and 90 which is a multiple of 10.
11, 101, 131, 151, 181, 191
Between the two numbers there are 49.
There are 43 natural numbers between 200 and 500 that are divisible by seven.
There are 16 numbers
11, 101, 131, 151, 181 and 191.
-200
I think the answer is: 66 numbers between 1 and 200 are divisible by 3. :) That's what I think the answer is.