Baker's Dozen

No.

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\).

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\).

Since given that y is a negative number (y<0), then \(|y| = -y\), hence \(-1-|y|=-1-(-y)=-1+y\).



Hope it helps.
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Hi Bunuel,

I was unsure how to do this problem and when I saw your explanation, I was wondering how you got ( 9 * 9 * 3C5 ) ? Not sure where the 9s came from.

Thanks,

Hi Brunel,
Great set of questions! Could you please explain the logic behind how you factorised this:
\(y=(3^5-3^2)^2*(5^7-5^4)^2=3^4*(3^3-1)^2*5^8*(5^3-1)^2\).
I am not too clear on how you pulled out factors of 3^4 and 5^8, and still have a 3^3 and a 5^3 left in the brackets.

Thanks!

Would really appreciate if someone can explain this bit "\sqrt{k} must complete the powers of 2 and 7 to even number,"
Why does it have to be powers of 2*7

Let me ask you: by what positive integer we should multiply \(126=2*3^2*7\) for the product to be the square of an integer?
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I did not get why 1/x is multiplied. Would you mind terribly, to explain?
Thanks

Responding to a pm:

Permutation is choosing and arranging - correct! But you need to know what you are arranging. Permutation (written as nPr) implies choose r things out of n and then arrange the r things among themselves. Say you select 3 chocolates out of 5 to give as prizes in a game and then arrange the 3 selected chocolates in three spots - first, second and third. The one who stands first, gets the first chocolate and so on. So if you use permutation here, you are selecting 3 spots where you will put 6 and then arranging them in 3! ways. What we actually need to do is select 3 spots out of 5 and put 6 in them. We have to then select 2 digits for the leftover 2 spots, wherever they are!

Hence we select the three spots in 5C3 ways and then choose a digit for the remaining two places in 9*9 ways.

Probability approach:
There are 5 spots ___ ___ ___ ___ ___

What is the probability that the number looks like 6N6N6? (1/10)*(9/10)*(1/10)*(9/10)*(1/10)
N is any number other than 6.

What is the probability that the number looks like N66N6? (9/10)*(1/10)*(1/10)*(9/10)*(1/10)

Similarly there are many other ways of making the password with three 6s. You need to add all some probabilities. The probability in each case will be \((1/10)^3 * (9/10)^2\). How many such distinct cases will be there? It depends on the number of ways in which you can arrange the 6s and Ns = 5!/3!*2! = 10
Total probability \(= 10*(1/10)^3 * (9/10)^2. = 810/10^5\)

Can I use permutation for the 2 non-6 digits? 5C3 is for choosing 3 spots to place 6s in 5 spots, for the remaining 2 spots one of 9 digits are used. Can this be expressed with '9C2' and something?

No. With 9C2, you are selecting 2 digits out of 9. There are two problems with it:
1. You are assuming that the two digits are distinct. Nothing stops the password from being 66622.
2. You are not considering that the two digits are for two distinct places. 66623 is different from 66632 but with 9C2, you are assuming that it is the same password.

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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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

Bunuel niks18






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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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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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.

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
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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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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

 

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


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