111, 121, 222, 212
111, 121, 222, 212
111, 121, 212, 222
There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.
111, 121, 131, 222, 212, 232, 333, 313, 323
676 of them.
111, 121, 222, 212
111, 121, 212, 222
There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.
111, 121, 131, 222, 212, 232, 333, 313, 323
Yuo can make only one combination of 30 digits using 30 digits.
676 of them.
There are 10 digits, but for a three digit number the first number cannot be a 0. Thus: there is a choice of 9 digits for the first (and last digit which must be the same), with 10 choices of digit for the second (middle) digit, making 9 × 10 = 90 such palindromic numbers.
Yes. But that is true only if the 100 digits do not include 0. Or, if 0 is included, then you consider "0n0" to be a three digit number. Most people would consider is to be a 2-digit number.
2578 5287 5782 8572
8643
531
Yes.