Baker's Dozen : Problem Solving (PS)

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Re: Baker's Dozen [#permalink]

19 Feb 2015, 18:22

Bunuel wrote:

4. What is the smallest positive integer $k$ such that $126*\sqrt{k}$ is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

$126=2*3^2*7$, so in order $126*\sqrt{k}$ to be a square of an integer $\sqrt{k}$ must complete the powers of 2 and 7 to even number, so the least value of $\sqrt{k}$ must equal to 2*7=14, which makes the leas value of $k$ equal to 14^2=196.

Answer: D.

Hi Bunuel,

Thanks for your great workouts. I have a question on this one though. If K= B (36), the number would become a square of a positive integer as well. For that matter any perfect square number will do -- say 4?!

$126*\sqrt{36}$ = 756. Why can't B be an answer. Your comment is highly appreciated.

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Re: Baker's Dozen [#permalink]

20 Feb 2015, 03:27

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falewman wrote:

Bunuel wrote:

4. What is the smallest positive integer $k$ such that $126*\sqrt{k}$ is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

$126=2*3^2*7$, so in order $126*\sqrt{k}$ to be a square of an integer $\sqrt{k}$ must complete the powers of 2 and 7 to even number, so the least value of $\sqrt{k}$ must equal to 2*7=14, which makes the leas value of $k$ equal to 14^2=196.

Answer: D.

Hi Bunuel,

Thanks for your great workouts. I have a question on this one though. If K= B (36), the number would become a square of a positive integer as well. For that matter any perfect square number will do -- say 4?!

$126*\sqrt{36}$ = 756. Why can't B be an answer. Your comment is highly appreciated.

$126*\sqrt{k}$ must be a prefect square. If k=36, $126*\sqrt{k}=756$, which is not a perfect square.

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Re: Baker's Dozen [#permalink]

11 Jan 2016, 06:33

Bunuel wrote:

SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is $9*9*C^3_5=810$: each out of two other digits (not 6) has 9 choices, thus we have 9*9 and $C^3_5$ is ways to choose which 3 digits will be 6's out of 5 digits we have.

$P=\frac{favorable}{total}=\frac{810}{10^5}$

Answer: B.

Hi Bunuel,

I have a doubt in 5C3 part in this problem.
I solved something like this:

Code will be something like this

6,6,6,A,B

this is like arranging 5 letters where 3 are exactly similar. It would give:

5! / 3! which is not equal to 5C3 .
Therefore I am getting a wrong answer.

Please correct my mistake (and why we are using Combination here while it looks like a case of Permutation).

Thanks in advance.

SR

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Re: Baker's Dozen [#permalink]

11 Jan 2016, 08:47

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solitaryreaper wrote:

Bunuel wrote:

SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is $9*9*C^3_5=810$: each out of two other digits (not 6) has 9 choices, thus we have 9*9 and $C^3_5$ is ways to choose which 3 digits will be 6's out of 5 digits we have.

$P=\frac{favorable}{total}=\frac{810}{10^5}$

Answer: B.

Hi solitaryreaper,

there are various Queries on this Q that why don't we use permutations and use combinations..
that is why 5!/3!2! instead of 5!/3!..

firstly, yes the Question is of permutation, but we still require to use combination formula..
WHY?
we have choosen three 6 digits and 2 digits as any of the remaining three..
answer is 9*9*5C3.... and not 9*9*5P3
because the permutations will be ok till the time we have two separate digits...
and it will be combinations the moment the other two digits are same....
so how can we use permutations even when digits are same, and that is what we are doing when we take other two digits as 9*9..

then how do we arrive at 9*9*5!/3!2!..
there are two ways we can take the remaining two digits..

1) three 6s and both other digits are different..
aaabc b and c can be chosen out of 9 digits
ways = 9C2*5!/3!=(9*8)/2! * 5!/3!...

2)three 6s and both other digits are same..
aaabb.. b can be selected in 9 ways
ways = 9*5!/3!2!..

total ways = add the two=9*8*5/3!2! + 9*5!/3!2!= 9*5!/3!2!* (8+1)=9*9*5!/3!2!=9*9*5C3..

I hope this clears the air around the use of permutation or combination in this Q..

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Re: Baker's Dozen [#permalink]

12 Jan 2016, 00:27

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solitaryreaper wrote:

Bunuel wrote:

SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is $9*9*C^3_5=810$: each out of two other digits (not 6) has 9 choices, thus we have 9*9 and $C^3_5$ is ways to choose which 3 digits will be 6's out of 5 digits we have.

$P=\frac{favorable}{total}=\frac{810}{10^5}$

Answer: B.

Hi Bunuel,

I have a doubt in 5C3 part in this problem.
I solved something like this:

Code will be something like this

6,6,6,A,B

this is like arranging 5 letters where 3 are exactly similar. It would give:

5! / 3! which is not equal to 5C3 .
Therefore I am getting a wrong answer.

Please correct my mistake (and why we are using Combination here while it looks like a case of Permutation).

Thanks in advance.

Now I am having a hard time to grasp how combination is giving the same answer. Can you please help me with the combination approach(how the problem translates to combination which is equivalent to the one the we got via permutation) ?

Moreover , how to decide when to use Permutation or Combination (and one that would lead to the solution in lesser time)

SR

Responding to a pm:

This is a question that haunts many - how do I know when to use permutation and when to use combination?
There is a simple solution - never use the permutation formula. The permutation formula finds very little direct use but leads to too many complications.

Always think in terms of selecting and arranging. For selecting r distinct elements out of n distinct elements, use nCr formula. Then arrange depending on whether the r elements selected need to be all distinct (r!) or some need to be same (a!/b!*c!) etc

Here, out of 9 digits you can select 2 in 9C2 ways and get a case such as 666AB. This will be arranged in 5!/3! ways.
Or you can select 1 digit out of 9 in 9C1 ways and get a case such as 666AA. This will be arranged in 5!/2!*3! ways.

9C2*5!/3! + 9C1*5!/2!*3! = 9*5!/3! * 9/2 = 810

Probability = 810/10^5

Note that you are using the Combinations formula only in this method too.

The other method using the combination formula just shows a different way of thinking.

You have 5 spots: _____ _____ _____ _____ _____

You choose any two spots out of these 5 in 5C2 ways.
For the first spot you choose, select a digit in 9 ways. For the second spot, select a digit in 9 ways.
In all remaining spots, just put 6.
This 5C2*9*9 directly gives you the total number of ways.
Note that 5C2 is the same as 5C3 (either you choose 2 spots for non 6 digits or you choose 3 spots for 6).

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Re: Baker's Dozen [#permalink]

13 Mar 2017, 09:01

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7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?
A. $60,000
B. $90,000
C. $100,000
D. $120,000
E. $180,000

Solution: https://gmatclub.com/forum/baker-s-dozen ... l#p1057509

Let F represents Fritz, L for Luis, A for Alfred and W for Werner

Fritz owns 2/3rd of the shares of the other three shareholders --> F = 2/3 (3,600,000 - F) ---> 3F = 7,200,000 - 2F ---> F = 1,440,000

Luis owns 3/7th of the shares of the other three shareholders --> L = 3/7 (3,600,000 - L) ---> 7L = 10,800,000 - 3L ---> L = 1,080,000

Alfred owns 4/11th of the shares of the other three shareholders --> A = 4/11 (3,600,000 - A) ---> 11A = 14,400,000 - 4A ---> A = 960,000

W= 3,600,000 -(F + L + A) ---> 3,600,000 - (1,440,000 + 1,080,000 + 960,000) = 120,000

Answer: D

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Re: Baker's Dozen [#permalink]

27 Mar 2017, 18:07

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Here's my solution for #2 (see visual below). It helps to use brackets!

Attachments

Screen Shot 2017-03-27 at 6.07.16 PM.png [ 182.8 KiB | Viewed 4340 times ]

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Re: Baker's Dozen [#permalink]

19 Dec 2017, 03:51

Bunuel niks18

Quote:

A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215

Quote:

Say 7 consecutive odd integers are: $x$, $x+2$, $x+4$, $x+6$, $x+8$, $x+10$, $x+12$.

I solved the same problem taking hint from number theory that a consecutive odd number is represented by:

2x+1 , 2x+3

My approach had slightly complex calculations but it ensured that my first no in series is always odd.

Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones)

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Re: Baker's Dozen [#permalink]

19 Dec 2017, 04:03

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adkikani wrote:

Bunuel niks18

Quote:

A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215

Quote:

Say 7 consecutive odd integers are: $x$, $x+2$, $x+4$, $x+6$, $x+8$, $x+10$, $x+12$.

I solved the same problem taking hint from number theory that a consecutive odd number is represented by:

2x+1 , 2x+3

My approach had slightly complex calculations but it ensured that my first no in series is always odd.

Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones)

In this question, as shown, you can take 7 consecutive odd integers to be $x$, $x+2$, $x+4$, $x+6$, $x+8$, $x+10$, and $x+12$. For some number properties questions, you should represent odd integers, as 2k + 1, 2k + 3, ... Here you can also take the integers, to be 2k + 1, 2k + 3, ... but it's not necessary.

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Re: Baker's Dozen [#permalink]

26 Mar 2018, 04:59

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nahid78 wrote:

Bunuel wrote:

SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.

# of passwords with three digit 6 is $9*9*C^3_5=810$: each out of two other digits (not 6) has 9 choices, thus we have 9*9 and $C^3_5$ is ways to choose which 3 digits will be 6's out of 5 digits we have.

$P=\frac{favorable}{total}=\frac{810}{10^5}$

Answer: B.

I know this is how we solve this question. But can anyone please clarify why i don't get the same result if i work differently ....

Suppose other two numbers are also same 6,6,6, and 1,1 or 6,6,6 and 2,2 so on... or other two are different. 6,6,6,1, and 2, or 6,6,6,2,and 4

You do get the same result in this case too.

If the other 2 numbers are same, number of ways = 9C1 * 5!/3!*2! = 90
If the other 2 numbers are different, number of ways = 9C2 * 5!/3! = 720

Total = 90+720 = 810

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Re: Baker's Dozen [#permalink]

21 Dec 2019, 08:27

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Bunuel wrote:

9. If x and y are negative numbers, what is the value of $\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}$?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y

Note that $\sqrt{a^2}=|a|$. Next, since $x<0$ and $y<0$ then $|x|=-x$ and $|y|=-y$.

So, $\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$

Answer: D.

Just Excellent Question Bunuel...

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Re: Baker's Dozen [#permalink]

22 Dec 2019, 04:35

Bunuel wrote:

8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215

Say 7 consecutive odd integers are: $x$, $x+2$, $x+4$, $x+6$, $x+8$, $x+10$, $x+12$.

Question: $x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?$

Given: $(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185$ --> $(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185$ --> $(5x+20)+20=-185$ --> $5x+20=-205$

Answer: D.

Hello Bunuel, I tried to solve it like this.. Can u please tell me where did I go wrong?

Let the smallest no. be 2n+1, so the numbers of the set will be :
2n+1, 2n+3, 2n+5, 2n+7, 2n+9, 2n+11, 2n+13

given that sum of largest 5 no.s in the set is -185.
so, 2n+5+2n+7+2n+9+2n+11+2n+13 = -185
solving, n=-23

so the no.s are -45,-47,-49,-51,-53,-55,-57

sum of smallest 5 are -49-51-53-55-57 = --265

Also, if we take into consideration that these are negative no.s, from the start we can write that 2n+1+2n+3+2n+5+2n+7+2n+9 = -185
solving n= -21
so the no.s are -41, -43, -45, -47, -49, -51, -53
sum of 5 smallest = -245

Really confused.. Please help.

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Re: Baker's Dozen [#permalink]

22 Dec 2019, 08:57

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dips1122 wrote:

Bunuel wrote:

8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is -185 what is the sum of the 5 smallest integers of set A?
A. -165
B. -175
C. -195
D. -205
E. -215

Say 7 consecutive odd integers are: $x$, $x+2$, $x+4$, $x+6$, $x+8$, $x+10$, $x+12$.

Question: $x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?$

Given: $(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185$ --> $(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185$ --> $(5x+20)+20=-185$ --> $5x+20=-205$

Answer: D.

Hello Bunuel, I tried to solve it like this.. Can u please tell me where did I go wrong?

Let the smallest no. be 2n+1, so the numbers of the set will be :
2n+1, 2n+3, 2n+5, 2n+7, 2n+9, 2n+11, 2n+13

given that sum of largest 5 no.s in the set is -185.
so, 2n+5+2n+7+2n+9+2n+11+2n+13 = -185
solving, n=-23

so the no.s are -45,-47,-49,-51,-53,-55,-57

sum of smallest 5 are -49-51-53-55-57 = --265

Also, if we take into consideration that these are negative no.s, from the start we can write that 2n+1+2n+3+2n+5+2n+7+2n+9 = -185
solving n= -21
so the no.s are -41, -43, -45, -47, -49, -51, -53
sum of 5 smallest = -245

Really confused.. Please help.

Since n = -21, then the seven consecutive numbers are: -45, -43, -41, -39, -37, -35, -33.

The sum of the five largest is -41 + (-39) + (-37) + (-35) + (-33) = -185.
The sum of the five smallest is -45 +(-43) +(-41) + (-39) + (-37) = -205.

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Re: Baker's Dozen [#permalink]

Updated on: 02 Feb 2021, 09:16

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Bunuel wrote:

I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

Total digits = 10 (from 0 to 9)
Of the 5 digits, we have 6 thrice.

Case 1: Let the other 2 digits be same:
Choose that digit 'n' in 9 ways and arrange the digits of the number 666nn in 5!/3!2! = 10 ways => Total ways = 90

Case 2: Let the other 2 digits be different:
Choose the digits 'n' in 9c2 = 36 ways and arrange the digits of the number 666mn in 5!/3! = 20 ways => Total ways = 720

Thus, total such numbers = 810
Total 5-digit numbers = 10 x 10 x 10 x 10 x 10 = 10000
=> Probability = 810/10000 - Answer B

Originally posted by sujoykrdatta on 02 Feb 2021, 08:45.
Last edited by sujoykrdatta on 02 Feb 2021, 09:16, edited 1 time in total.

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Re: Baker's Dozen [#permalink]

02 Feb 2021, 09:05

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Bunuel wrote:

I'm posting the next set of medium/hard PS questions. I'll post OA's with detailed explanations after some discussion. Please, post your solutions along with the answers. Good luck!
13. If $x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}$, what is the product of the tens and the units digits of $\frac{x}{(8!)^3}-39$?
A. 0
B. 6
C. 7
D. 12
E. 14

Solution: https://gmatclub.com/forum/baker-s-dozen ... l#p1057520

x = [(8!)^10−(8!)^6] / [(8!)^5−(8!)^3] = {(8!)^6 x [(8!)^4 − 1]} / {(8!)^3 x [(8!)^2 − 1]} = (8!)^3 x [(8!)^2 + 1]

Thus, x/(8!)^3 - 39 = (8!)^2 − 38

8! ends in 0; thus, (8!)^2 ends in 00 i.e. it is a multiple of 100

Thus, the last 2 digits of (8!)^2 − 38 are 62 => Product = 12

Answer D

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Re: Baker's Dozen [#permalink]

12 Aug 2023, 08:39

Bunuel wrote:

SOLUTIONS:

1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?

A. 860/90,000
B. 810/100,000
C. 858/100,000
D. 860/100,000
E. 1530/100,000

The total number of 5-digit codes is $10^5$. It's important to note that it's not $9*10^4$, as the first digit can be zero in a password.

The number of passwords with three digits as 6 can be calculated as $9*9*C^3_5 = 810$. We have 9 choices for each of the two remaining digits (not 6), resulting in $9*9$. The term $C^3_5$ represents the number of ways to choose the positions for the 6s among the five digits (essentially deciding which three out of ***** will be 6s).

The probability is therefore, $P=\frac{favorable}{total}=\frac{810}{10^5}$.

Answer: B

why we didn't multiply it with 2! ? two non-six numbers can be arranged in 2 ways...it is a password so we have to go for arrangement...right? please someone help me to understand

regards

pudu

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Baker's Dozen [#permalink]

05 Mar 2024, 04:22

Bunuel wrote:

13. If $x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}$, what is the product of the tens and the units digits of $\frac{x}{(8!)^3}-39$?
A. 0
B. 6
C. 7
D. 12
E. 14

Apply $a^2-b^2=(a-b)(a+b)$: $x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}=\frac{((8!)^{5}-(8!)^3)((8!)^{5}+(8!)^3)}{(8!)^{5}-(8!)^3}=(8!)^{5}+(8!)^3$.

Next, $\frac{x}{(8!)^3}-39=\frac{(8!)^{5}+(8!)^3}{(8!)^3}=\frac{(8!)^{5}}{(8!)^3}+\frac{(8!)^{3}}{(8!)^3}-39=(8!)^2+1-39=(8!)^2-38$.

Now, since $8!$ has 2 and 5 as its multiples, then it will have 0 as the units digit, so $(8!)^2$ will have two zeros in the end, which means that $(8!)^2-38$ will have 00-38=62 as the last digits: 6*2=12.

Answer: D.

First of all thanks fort he series
I approached the problem in a diff way
Replace 8! with A for simpler maths -> and write faster

$x=\frac{(A)^{10}-(A)^6}{(A)^{5}-(A)^3}\frac$.
We extract a^3
$x=\frac{(A)^{3}*(A)^{3}*((A)^{4}-1)}{(A)^{3}*(A)^{2}-1}\frac$.
Now on the top, we have:
$a^2-b^2=(a-b)(a+b)$
$x=\frac{(A)^{3}*(A)^{3}*((A)^{2}-1)*((A)^{2}+1)}{(A)^{3}*((A)^{2}-1)}\frac$.
So we can simplify
$x={(A)^{3}*((A)^{2}+1)}$.
as we have x/8!^3
${((A)^{2}+1)}-39$.
then we reach
$(8!)^2-38$
and we can proceed on the same way (2*5)^2 is 00 so 62 -> 6*2 = 12

gmatclubot

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