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.
Oh, dude, so like, a palindrome is a number that reads the same forwards and backwards, right? And it's between 200 and 300? So, if we think about it, the only 3-digit palindrome in that range is 202. So, the difference between the two 3-digit numbers must be 202. Easy peasy, lemon squeezy!
There are 900 three-digit numbers, ranging from 100 to 999.
266, 267, 268
111,112,121,211,122,212,221,222,113,131,311,133,313,331,333,223,232,322,233,323,332,123, 132, 213, 231, 312, 321
the answer is 135
I dont know and i really need the answer for very necessesary reason..
2 is prime, its only factors are one and itself.
12
0.12 Or 102 if you do not want to include non integers.
The smallest 3 digit counting number less than 1000?? 100
196, 198, 202, 204
4*9 = 36 so 360 is one possible answer.
it is int n=123; printf ("%d", (n/10)%10);