To solve the problem, we need to find all 3-digit palindromes such that the product of their digits has 1 as the unit digit.
A 3-digit palindrome can be represented as aba , where a and b are digits, and a≠0 because it's a 3-digit number.
The product of the digits in the palindrome is a×b×a = a^2 ×b.
For the product
a^2 ×b to have 1 as the unit digit, we need to find suitable values of a and b.
Let's analyze:
The unit digit of a^2 must multiply with b to give a unit digit of 1.
Considering possible unit digits:
If a ends in 1: a^2 ends in 1.
If a ends in 2: a^2 ends in 4.
If a ends in 3: a^2 ends in 9.
If a ends in 4: a^2 ends in 6.
If a ends in 5: a^2 ends in 5.
If a ends in 6: a^2 ends in 6.
If a ends in 7: a^2 ends in 9.
If a ends in 8: a^2 ends in 4.
If a ends in 9: a^2 ends in 1.
If a ends in 0: a^2 ends in 0 (not possible since a≠0).
We see that a ending in 1 or 9 will make
a2 end in 1, which is necessary for the product to end in 1 when multiplied by b.
Next, we need b such that
1×b ends in 1. The only digit
b can be is 1 because 1×1=1.
So, we only need to consider palindromes of the form a1a .
The palindromes fitting the criteria are:
If a=1: the number is 111.
If a=9: the number is 919.
So, the minimum possible value of the tens digit of any of these numbers is 1, and the maximum possible value is 1.
Therefore, the answers are:
Minimum: 1
Maximum: 1