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# is a 3 digit palindrome
Product of 3 digits ends with 1
We can eliminate all even digits and 5
The only digits we can use are 1,3,7,9

The numbers that satisfy all four criteria: 797, 919, 393

Max value of tens digit = 9
Min value of tens digit = 1
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Bunuel
A palindrome is a number that reads the same when read from either end. For example, 12321 is a palindrome. A set consists of all the 3-digit palindromes whose product of digits has 1 as the unit digit. Select for Minimum the minimum possible value of the tens digit of any of the numbers in the set, and select for Maximum the maximum possible value of the tens digit of any of the numbers in the set. Make only two selections, one in each column.

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Lowest - 111 , maximum - 797

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A palindrome is a number that reads the same when read from either end. For example, 12321 is a palindrome. A set consists of all the 3-digit palindromes whose product of digits has 1 as the unit digit. Select for Minimum the minimum possible value of the tens digit of any of the numbers in the set, and select for Maximum the maximum possible value of the tens digit of any of the numbers in the set. Make only two selections, one in each column.

3-digit palindromes whose product of digits has 1 as the unit digit: 111, 393, 797
Minimum 1: => 111
Maximum: 9 => 797­
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Three Digit palindrome number can be of the form xxx or xyx

Now x^3 last digit =1
or x{^2}*y last digit =1
We need to find minimum x or y
Minimum x=1, when Number = 111

Maximum y=9 when Number 393

Imo 1-2, 2-6
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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
­
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Catman
Three Digit palindrome number can be of the form xxx or xyx

Now x^3 last digit =1
or x{^2}*y last digit =1
We need to find minimum x or y
Minimum x=1, when Number = 111

Maximum y=9 when Number 393

Imo 1-2, 2-6
­Bunuel I have not received Kudos for this explanation.Can you please check? Thanks.­
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Catman

Catman
Three Digit palindrome number can be of the form xxx or xyx

Now x^3 last digit =1
or x{^2}*y last digit =1
We need to find minimum x or y
Minimum x=1, when Number = 111

Maximum y=9 when Number 393

Imo 1-2, 2-6
­Bunuel I have not received Kudos for this explanation.Can you please check? Thanks.­
­Kudos given. Thank you for reporting.
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