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GMAT Club Olympics 2024 (Day 10): A palindrome is a number that reads

1 2 3 4

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19 Jul 2024, 21:08

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A 3-digit palindrome has the form aba, where a and b are digits, and a cannot be zero.The number is 101a+10b
Since $a^2$ should multiply with b to give a unit digit of 1, possible values for a (since a ranges from 1 to 9) must make $a^2$ yield a unit digit such that when multiplied by b, results in a unit digit of 1.
a can be 1,3,7,9 since only these values have $a^2$ ending in 1 or 9, which can yield 1 when multiplied by some b.
For a=1, b should be 1.
For a=3, $a^2=9$, For this case b should be 9 so that 3*9*3=81.
So, we can see that the minimum tens digit is 1 and maximum is 9.

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19 Jul 2024, 21:09

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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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19 Jul 2024, 21:54

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Quote:

Let make a list of all plaindromes with the given restrictions:-

111
393
797
919

We can see that maximum tens' digit = 9, and minimum ten's digit = 1

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19 Jul 2024, 22:32

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We need to find 2 numbers that when multiplying give 1 in its units digits, which is an odd number, so we discard even numbers because we any number multiplied by an even number gives an even number. Also we need to discard 5 because any number multiplied by 5 give 5 or zero in the unit digit.
So the numbers we have are 1,3,7,9
so possible palindromes are
111
333
777
999
131
313
737
717
939
919
When multiplying them

1*1*1=1
3*3*3=27
7*7*7=343
9*9*9=729
1*3*1=3
1*7*1=7
1*9*1=9
3*1*3=9
7*3*7=347
7*1*7=21
7*9*7=441
9*3*9=243
9*1*9=81
3*7*3=63
3*9*3=81
9*7*9=567

1*1*1=1
9*1*9=81
7*9*7=441
3*9*3=81

So the minimum number for tent digit is 1 and the max number is 9
Minimum 1
Maximum 9

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19 Jul 2024, 22:59

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The numbers are 3-digit palindromes.
The product of digits has 1 as the unit digit.

Let's check the options:

For minimum,
0 - 0 cant be as the product will become 0,
1 - If the tens digit is 1, digit is in the form x1x.
x can be 9 as 9*9 = 81 will result in 1 as the unit digit of the product.
So, 1 is possible.

For maximum,
9 - If the tens digit is 9, digit is in the form x9x.
x can be 3 as 3*3 = 9 will result in 1 as the unit digit of the product. (3*9*3 = 81)
So, 9 is possible.

Minimum: 1
Maximum: 9

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20 Jul 2024, 00:01

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Set-A Possible 3-digit palindromes options will be - (111), (797) & (919) when multiplication of all digits has 1 as the unit digit.

Tens digit =1,9

Hence -

Minimum- the minimum possible value of the tens digit of any of the numbers in the set - 1

Maximum- the maximum possible value of the tens digit of any of the numbers in the set - 9

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20 Jul 2024, 00:09

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Given - 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.

To find - 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.

A set consists of all the 3-digit palindromes will be in "$aba$" form
and their product of the digits has 1 as the unit digit.
Means mathematically it will be like
$a^2*b = xx1$
now we need to find max and min value for b.
min of b can't be zero because it will make product 0
So,
range of b will be from 1 to 9
so lets check.
we want final product to have 1 as unit digit,
and that can be happen with 3 digit numbers for $bmin = 1$ when a^2 ends with the unit digit place as 1, and this will happen when a is 1 or 9.
Now take a look at following digits formation.
when a=1 and b=1, aba = 111 = 1*1*1 = 1
and when a=9 and b=1, aba = 919 = 9*1*9 = 81

as question asked us for any of number in the set. Hence for $bmin = 1$ we have two possible 3 digit formations 111 and 919.

Now lets check for $bmax$ I will start with max digit 9. $bmax = 9$
So now for product of $a^2*b$ to be ended in $"xx1"$ with the value of $bmax$ as 9.
we will need $a^2$ to end in unit digit $xx9$ and further we know when we multiply that value of $a^2$ with 9 will get number that will ends with unit digit 1.
So the possible values for $a^2$ which will lead us to number in the form of $xx9$ are 3 and 7.
Now take a look at following digits formation.
when a=3 and b=9, aba = 393 = 3*9*3 = 81
and when a=7 and b=9, aba = 797 = 7*9*7 = 441

as question asked us for any of number in the set. Hence for $bmax = 9$ we have two possible 3 digit formation 393 and 797.

Minimum $bmin = 1$ Option B
Maximum $bmax = 9$ Option F

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20 Jul 2024, 00:28

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Min value =1 because 111 gives product of 1 in UD
Max value =9 because 393 gives product of 1 in UD

Ans B, F

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20 Jul 2024, 00:44

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The product of digits should have 1 at the Units place.

So smallest number would be 111. 101 cannot be the smallest one because the product of digits is 0.
Hence, minimum is 1

Let's start checking with 9 for maximum
_9_ : find in this structure that will give 1 at units digit. 393 = 3*9*3 = 81. So maximum is 9

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20 Jul 2024, 00:56

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All the potential palindromes will be:
1_1, 2_2, 3_3, 4_4, 5_5, 6_6, 7_7, 8_8, 9_9

To get 1 at the end of the product, we have to eliminate all even numbers (because they only end in evens) and 5, vecause it'll only end in 5 or 0.
So, we're left with:

1_1: 1 gives 1 at the end when multiplied by 1 only, so 111
3_3: 3*3=9 gives 1 at the end when multiplied by 9 only, so 393
7_7: 7*7=49 gives 1 at the end when multiplied by 9 only, so 797
9_9: 9*9=81 gives 1 at the end when multiplied by 1 only, so 919

As we can see, the maximum value for the middle digit is 9, and the minimum is 1.

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20 Jul 2024, 00:58

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Minimum =919 (product is 81)
Maximum =393 (product is 81)

Posted from my mobile device

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20 Jul 2024, 01:06

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Product of 3 and 7, 9 and 9 gives the unit digit 1.
Possible sets are:393,797,919
Min:1; Max:9

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20 Jul 2024, 01:29

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Here we are.
At the finish.
We need a miracle to win this competition.
But we gave whatever we could.
For the last time, here is my explanation:

Glance - the Question&Answer Choices:
Question: We are dealing here with a counting and kind of 'find the pattern' question.
Answer Choices: They do not help us guess or estimate because they are close to each other and what we are looking for are jointly consistent.

Rephrase - Reading and Understanding the question:
Given: Okay, let’s collect what we have here:
palindrome = "12321", we are dealing with 3 number palindrome.
Our palindrome = 'XYX'
X*Y*X = ___1 (1 at the end/unit digit) => X^2*Y
[?]: Maxmimum and minimum we can come up of Y.

Solve:
What numbers we can multiply and get 1 at the end?
9 * 9 = 81 , 7 * 3 = 21 , 1 * 1 = 1
Alright now we know we need to multiply 3 number by the equation: X^2*Y
so the possibilities palindromes we have in order to get 1 are:
111 , 919 , 393 , 797.
What is the minimum and maximum value of y?
Clearly, The answers are 1 and 9 (respectively).

THE END
I hope you liked the explanation; I did my best here.
Please let me know if you have any further questions about this question or my explanation.

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20 Jul 2024, 01:30

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Bunuel

Given options: 0 1 3 5 7 9

Minimum possible digit is 0
But, if any of the digits is zero, then the product of the digits becomes 0 and unit digit cannot be 1.
Hence, none of the digits are 0.

Next possible minimum digit is 1
we can make a palindrome with 1 in tens place that satisfies the conditions = 111
111 is a palindrome with product having 1 at the units digit

Minimum = 1

Maximum possible digit = 9
palindrome following the given conditions = 393

Maximum = 9

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20 Jul 2024, 01:31

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Since the set consists of all the 3-digit palindromes whose product of digits has 1 as the unit digit, let's look at those numbers

____ ____ ____

Because it is a palindrome, units digit and hundreds digit will be same

1 1
____ ____ ____

The only option for the tens digit will be 1 in order for the product of the digits to be 1 then it will be 111.

Likewise, the other 3-digit palindromes would be 393, 797, 919.

Therefore, the minimum possible value of the tens digit of any of the numbers in the set is 1
and the maximum possible value of the tens digit of any of the numbers in the set is 9.

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20 Jul 2024, 02:30

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Answer: Min: 1 Max: 9

3-digit palindrome whose product of digits has 1 as its unit's digit.

111 - Palindrome number. Product of digits: 1. (units digit is 1)
Ten's digit of 111 is 1.

393 - Palindrome number. Product of digits: 81. (units digit is 1)
Ten's digit of 393 is 9.

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20 Jul 2024, 02:43

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The answer is 1 for minimum and 9 for maximum
Since the hundredth digit and ones digit will have the same number to qualify as a palindrome we need to find a number for the tens digit which will ensure that the in the product of the three digits, theres 1 in the ones digit
This can be possible if we take 3x9X3 =81 for maximum value and 9x1x9=81 for the minimum tens value.

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20 Jul 2024, 02:44

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We need to find 3 digits palindrome whose product of digits has $1$ as the unit digit.
And from these palindormes we need to find the minimum possible value of the tens digit of any of the numbers in the set, and the maximum possible value of the tens digit of any of the numbers in the set.

We can solve the question by logic.

For minimum,
What's the minimum value among options? $1$
Then assume 3 digits number is $x1x$.
Can we find any value of x such that multiplying it's square (square because we will be multiplying $x * x * 1$) with 1 will give us value with 1 as its unit digit?
Yes we can.
Remember $1 * 1 = 1$
Then if we take $x = 1$,
We get 3 digits number $111$.
Then product of digits = $1 * 1 * 1 = 1$
Is it palindrome? Yes. Is the product of digits has 1 as the unit digit? Yes.

Select 1 for minimum

For maximum,
Same process but start with maximum value.
What's the maximum value among options? $9$
Assume 3 digits number is $x9x$.
Can we find any value of x such that multiplying it's square (square because we will be multiplying $x * x * 9$) with 9 will give us value with $1$ as its unit digit?
Yes we can. $3$
Remember $3 * 3 = 9$ and $ 9 * 9 = 81 $
Then if we take x = $3$,
We get 3 digits number $393$.
Then product of digits = $3*9*3 = 81$
Is it palindrome? Yes. Is the product of digits has 1 as the unit digit? Yes.

Select 9 for maximum

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20 Jul 2024, 03:31

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Product with unit digit one are 21 and 81 and 1 itself.

81 is generated by 9x 9 = 393
21 is generated by7X 3 = a palindrome can be formed out of this
01 is generated by1x1 = 111

Hence tens digit can vary from 1 to 9

So minimum is 1 and maximum is 9 .