Baker's Dozen

Responding to a pm:


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That's a tough 700+ problem.

We are interested 666XX numbers. Now, two this can be arranged in 10 ways;
666XX;
66X6X;
6X66X;
X666X;

66XX6;
6X6X6;
X66X6;

6XX66;
XX666;

X6X66.

So, basically with \(C^3_5=10\) we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: \(C^2_510\) choosing which 2 places out of 5 will be occupied by 6's.

Next, each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0.

Total \(9*9*C^3_5=810\).

Hope it's clear.
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Consider this: \((ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2\), so \((3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2\).

Hope it helps.
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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

You figure it out by knowing that there is a 5 as a factor in the final number.

example 8! = 8*7*6*5...*1

Now if u know how many fives are there you will have that number of zeros in the end because at least that many number of 2 will there for sure ...

Now IF u want to figure out how may fives are there in (which is equivalent to finding how many zeros in the end of the number )
then simply keep dividing the factorial by 5 and adding it to the result until 5^x exceeds the factorial, i know i have used very confusing language but a example will simplify things for sure

how many zeros at the end of 312!
answer :- 312!/5 + 312!/5^2 + 312!/5^3 = 62 + 12 + 2 (now note i stopped at 5^3, because 5^4 =25*25=625 which is more that 312 ) hence :- 62 + 12 + 2 =76 number of 5s in the factorial and same number of zeros in the end because these many number of 2s will obviously be there in factorial and will form 10s .

IF you are still confused I would suggest you to download Bunuel's math's notes and even if you are not confused i would still suggest you to do the same ,

x ;x+2;x+2*2;x+2*3....x+2*6

since the set has consecutive odd integers, then mean =median

mean of 5 largest integers=-185/5= -37
median of 5 largest integers is the 3d of the largest or the 5th of the whole set A= x+2*4
x+8=-37 ;x=-45

median of the first 5 small integers =-45+2*2=-41


the sum of the first small odd integers=number of integers*mean (or median)=5*median=5*(-41)=-205
answ is
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I am not so certain whether I am right but I would go with D in #9.

IF x<0 and y<0 then sqrt(x^2)/x=|x|/x, then -x/x=-1; because there is a minus in front of x, than -(-x) should be positive.

sqrt two negative y gives square root of y^2, than because y<1, y is negative as well. -y.

From the expression we've got -1+y, thus D should be a correct answer.

Hi Bunuel,
Please help me in understanding the below text. I think I am lacking this concept.I always try to solve by taking LCM in order to get rid of the fractions. However, I found your approach short n simple. If you could shed some light on this, would be really helpful

Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares

Thanks
H

Since we are talking about a 5-digit password and not a 5-digit number, then it can start with zero. For example a password can be 00123 or 00001.

Hope it's clear.
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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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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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In regards to question 13, how do you know that (8!)^2 will have two 0's at the end?

If \(x<0\) then \(|x|=-x\) and \(\frac{|x|}{x}=\frac{-x}{x}=-1\), (\(-\frac{x}{x}=-1\) no matter whether \(x\) is negative or positive).
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Hi Bunuel,

I didn't quite understand this ..

X and Y are odd then how can |x| = -x it has to be +ive.
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x and y are negative numbers, not odd numbers.

If \(x\geq{0}\) then \(|x|=x\);
If \(x<{0}\) then \(|x|=-x\). So if \(x\) is negative then \(|x|=-x=-negative=positive\). For example, if \(x=-2\) then \(|x|=|-2|=2=-x\).

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
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I tried feagure out this way first but thought to calculate little differently-

Since there are a toatl of 12 Marbels and after picking the required # marbels we are left with 4 marbels in the Jar- so our task is determine out of these 4 in how many ways we can pick R, B marbels so that there exists at least 1 marbel of each type- (R-1,B-3), (R-2,B-2),(R-3,B-1)

So the required # ways = \(7C1*5C3 + 7C2*5C2 + 7C3*5C1\)
= \((7 * 10) + (21 * 10)+ (35 * 5)\)
= \(70 + 210 + 175\)
= 455 (D)

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Responding to a pm:

From the options, it is obvious that the marbles are considered unique i.e. the red marbles are all different, say they are numbered, and the blue marbles are all different. The problem with your approach is that when you leave 1 red and 1 blue, you don't provide for this.
Also, once you account for the uniqueness, there will be double counting - when you leave Red1 and Blue1 and then choose 8 out of the remaining 10 and say you are left with Red2 and Red3 too, this will be counted as different from when you leave Red2 and Blue1 and then choose 8 out of the remaining 10 and you are left with Red1 and Red3. These two cases are actually the same.

Rule of thumb is that 'at least one' cases are best done using the complement approach which is 'Total - None' (as shown by Bunuel in the solution).
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When m is divided by n, the remainder can be anything from 0 to (n-1).
Take examples:

When m is divided by 8, the remainder can 0 or 1 or 2 till 7. Can the remainder be 8? No because that means the number is divisible by 8.

Say you divide 20 by 8, you get remainder 4. You divide 21 by 8, you get remainder 5. You divide 22 by 8, you get remainder 6. You divide 23 by 8, you get remainder 7. You divide 24 by 8, you get remainder 0, isn't it? The remainder will not be 8 - it will be 0.

Hence if you know that the remainder when you divide m by n is 7, you can say that n must be at least 8.

Similarly, when a number is divided by z and remainder is 8, z must be at least 9.
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