Quote:
A palindrome is a number that reads the same forward and backward. For example, 3663 and 23232 are palindromes. If 7-digit palindromes are formed using one or more of the digits 5, 6 and 0, how many such palindromes are possible?
We know that our palindrome consists of 7 digits and each digit can be 0, 5 or 6. The task is to find the number of possible palindromes.
The important condition is that palindromes read the same forward and backward i.e. the second part of the number in this case mirrors the first part and in our case. As we have 7 digits,
first 3 will be mirrored by latter 3 and the 4th digit will be a unique one.
It means that latter 3 digits (5th, 6th and 7th) are dependent on the values of first 3 digits (1st, 2nd and 3rd). Taking it into account,
the latter 3 digits do not play role in composing unique numbers as they in any case are the same as previous 3.
Also note that the first digit cannot be 0, otherwise there will be less than 7 digits. Thus, first digit can be only 5 or 6 i.e. has 2 values.
2nd digit can have all three values: 0, 5 and 6.
The same applies to 3rd and 4th digits.
As it was found earlier, latter three digits are dependent on the first three, and for this reason they do not add up to the total number of palindromes.
Let us now count the possible number of palindromes:
\(2*3*3*3 = 2*3^3 = 2* 27 = 54\)
Answer:
B