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Re: Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
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filippocarta01 wrote:
Let s be the set of all 6-digit numbers of the form \(54x,y12\), where each of \(x\) and \(y\) can be any of the digits 3, 5, 8, and x = y is allowed. What is the probability that a randomly chosen number from set S is divisible by 8?

(A) \(0\)

(B) \(\frac{1}{9}\)

(C) \(\frac{2}{9}\)

(D) \(\frac{1}{3}\)

(E) \(\frac{2}{3}\)


Let me discuss on certain observations/analysis, which could help you in arriving at answer faster.

Divisibility rule of 8
If last 3 digits are divisible then the entire number is divisible by 8.

Why?
Because any number with last three digits zero is always divisible by 8. For example, 123000 or 876000 or 111000. (1000 is 8*125 and any of the numbers can be written as 111*1000 or 111*8*125).
We utilise the above fact for our divisibility rule.
Any number abcdef can be written as abc000+d00+ef. Here abc000 is divisible by 8.
So, we check def for divisibility by 8.

Some more facts
When you divide 100, 300, 500, 700 or 900 with 8, it will leave a remainder 4, so the last two digits should also give 4 as remainder.
For example 128 = 100+28 = 100+24+4
When you divide 200, 400, 600, 800 or 000 with 8, it will leave a remainder 0, so the last two digits should also give 0 as remainder.
For example 224 = 200+24

Back to question
Last 2 digits are 12, so the remainder is 4, implying that the number should have odd hundreds digit=> 3 and 5 are the possibility for y
As the thousands digit does not matter, y could be any of the three digits.
Total ways number is divisible by 8 is 3*2 or 6
Total ways to select x and y is 3*3 or 9
Hence the probability is \(\frac{6}{9}\) or \(\frac{2}{3}\)

E
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Re: Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
chetan2u wrote:
filippocarta01 wrote:
Let s be the set of all 6-digit numbers of the form \(54x,y12\), where each of \(x\) and \(y\) can be any of the digits 3, 5, 8, and x = y is allowed. What is the probability that a randomly chosen number from set S is divisible by 8?

(A) \(0\)

(B) \(\frac{1}{9}\)

(C) \(\frac{2}{9}\)

(D) \(\frac{1}{3}\)

(E) \(\frac{2}{3}\)

Let me discuss on certain observations/analysis, which could help you in arriving at answer faster.

Divisibility rule of 8
If last 3 digits are divisible then the entire number is divisible by 8.

Why?
Because any number with last three digits zero is always divisible by 8. For example, 123000 or 876000 or 111000. (1000 is 8*125 and any of the numbers can be written as 111*1000 or 111*8*125).
We utilise the above fact for our divisibility rule.
Any number abcdef can be written as abc000+d00+ef. Here abc000 is divisible by 8.
So, we check def for divisibility by 8.

Some more facts
When you divide 100, 300, 500, 700 or 900 with 8, it will leave a remainder 4, so the last two digits should also give 4 as remainder.
For example 128 = 100+28 = 100+24+4
When you divide 200, 400, 600, 800 or 000 with 8, it will leave a remainder 0, so the last two digits should also give 0 as remainder.
For example 224 = 200+24

Back to question
Last 2 digits are 12, so the remainder is 4, implying that the number should have odd hundreds digit=> 3 and 5 are the possibility for y
As the thousands digit does not matter, y could be any of the three digits.
Total ways number is divisible by 8 is 3*2 or 6
Total ways to select x and y is 3*3 or 9
Hence the probability is \(\frac{6}{9}\) or \(\frac{2}{3}\)

E

Is there any way of doing it without recalling the rule of "divisibility by 8"? chetan2u and KarishmaB
I was trying to find a pattern and then seggregated the 6 digit number into 3 different patterns, 312, 512, 812. Then I tried diving the entire 6 digit number by 8 and simply went with it. Took time and I wasn't sure I doing the right thing and this strategy may not work for all questions of this type. Thank you.­
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Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
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Engineer1

Essentially what you are using is divisibility and operations on remainders. 

When I have a number \(54x,y12\), I can write it as \(54x000 + y12 = 54x * 1000 + y12\)

1000 is divisible by 8 so we need y12 to be divisible by 8 (that is why the divisibility by 8 rule is what it is. I discuss the reason behind each divisibility rule in my content. Check it out since it is free for all today under the Super Sundays program on anaprep. If you undertsand from where the rules come, you don't need to "remember" them. You will just "know" them automatically. )

312 and 512 are divisble by 8 but 812 is not.
Hence answer will be 2/3

Answer (E)­
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Re: Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
Expert Reply
 
filippocarta01 wrote:
Let s be the set of all 6-digit numbers of the form \(54x,y12\), where each of \(x\) and \(y\) can be any of the digits 3, 5, 8, and x = y is allowed. What is the probability that a randomly chosen number from set S is divisible by 8?

(A) \(0\)

(B) \(\frac{1}{9}\)

(C) \(\frac{2}{9}\)

(D) \(\frac{1}{3}\)

(E) \(\frac{2}{3}\)

­
Divisibility of 8: Last 3 digits of the number should be divisible by 8 or should be 000 e.g.123848 is divisible by 8 because 848 is divisible by 8


Now x and y may be {3, 5, 8}
and \(54x,y12\) ,ust be divisible by 8
i.e. x can be anything as it's not the part of last three digits of the number so x may be chosen in 3 ways

Now, if y = 3, 54x,y12 = 54x,312 and 312 is divisible by 8
if y = 5, 54x,y12 = 54x,512 and 512 is divisible by 8
if y = 8, 54x,y12 = 54x,812 and 812 is NOT divisible by 8
so we have two favorable cases for value of y

Total favorable cases = cases of x * cases of y = 3*2 = 6

Total Outcomes = 3*3 (both x and y have 3 choices each) = 9

Probability = 6/9 = 2/3

Answer: Option E

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Re: Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
 
KarishmaB wrote:
Engineer1

Essentially what you are using is divisibility and operations on remainders. 

When I have a number \(54x,y12\), I can write it as \(54x000 + y12 = 54x * 1000 + y12\)

1000 is divisible by 8 so we need y12 to be divisible by 8 (that is why the divisibility by 8 rule is what it is. I discuss the reason behind each divisibility rule in my content. Check it out since it is free for all today under the Super Sundays program on anaprep. If you undertsand from where the rules come, you don't need to "remember" them. You will just "know" them automatically. )

312 and 512 are divisble by 8 but 812 is not.
Hence answer will be 2/3

Answer (E)­

­
Thank you, Karishma. I follow ANAPrep on Youtube and will check. Is your video about the divisibilty rules and would you mind sharing a link?
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Re: Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
Expert Reply
Engineer1 wrote:
KarishmaB wrote:
Engineer1

Essentially what you are using is divisibility and operations on remainders. 

When I have a number \(54x,y12\), I can write it as \(54x000 + y12 = 54x * 1000 + y12\)

1000 is divisible by 8 so we need y12 to be divisible by 8 (that is why the divisibility by 8 rule is what it is. I discuss the reason behind each divisibility rule in my content. Check it out since it is free for all today under the Super Sundays program on anaprep. If you undertsand from where the rules come, you don't need to "remember" them. You will just "know" them automatically. )

312 and 512 are divisble by 8 but 812 is not.
Hence answer will be 2/3

Answer (E)­

­
Thank you, Karishma. I follow ANAPrep on Youtube and will check. Is your video about the divisibilty rules and would you mind sharing a link?

­There is an entire chapter on divisibility rules in my content (video + book  + practice questions). The video is not on YT but today my content is available to everyone (till 10:00 AM IST Monday) for free. You can access is through the "Super Sundays" bundle by registering on https://anaprep.com
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Re: Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
gmatophobia wrote:
filippocarta01 wrote:
Let s be the set of all 6-digit numbers of the form \(54x,y12\), where each of \(x\) and \(y\) can be any of the digits 3, 5, 8, and x = y is allowed. What is the probability that a randomly chosen number from set S is divisible by 8?

(A) \(0\)

(B) \(\frac{1}{9}\)

(C) \(\frac{2}{9}\)

(D) \(\frac{1}{3}\)

(E) \(\frac{2}{3}\)

Divisibility by 8 Rule: To check if a number is divisible by 8, simply see if the last three digits are divisible by 8. If they are, the entire number is divisible by 8.

In this case, \(y12\) must be divisible by 8 for \(54xy12\) to be divisible by 8.

Possible values of y = {3,5,8}

312 → Divisible by 8
512 → Divisible by 8
812 → Not Divisible by 8

Hence, \(y\) can be filled in two ways (i.e. by 3 & 5). \(x\) can be filled in \(3\) ways (i.e. using 3, 5, and 8).

Favorable cases = 3 * 2 = 6

The positions denoted by x and y can be filled in three ways i.e. we can have three possible values in each position denoted by x and y.

Total cases = 3 * 3 = 9

Required Probability = \(\frac{6}{9} = \frac{2}{3\\
}\\
\)

Option E

­i missed this question because I got favorable cases = 2 and total cases = 6 for 1/3. Why are we multiplying these results?

 
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Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
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Fish181 wrote:
­i missed this question because I got favorable cases = 2 and total cases = 6 for 1/3. Why are we multiplying these results

Fish181

Favorable cases = 6

\(y\) can be filled in 2 ways and \(x\) can be filled in 3 ways

Hence, favourable cases = 2 * 3 = 6

Here are possible values of xy

33
53
83
35
55
85

Total cases = 9, because \(x\) can be filled in 3 ways and \(y\) can be filled in 3 ways

Here are all the possible values of xy

33
53
83
35
55
85
38
58
88

Hence required probability = \(\frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{6}{9} = \frac{2}{3}\)

 ­
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Re: Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
gmatophobia wrote:
Fish181 wrote:
­i missed this question because I got favorable cases = 2 and total cases = 6 for 1/3. Why are we multiplying these results

Fish181

Favorable cases = 6

\(y\) can be filled in 2 ways and \(x\) can be filled in 3 ways

Hence, favourable cases = 2 * 3 = 6

Here are possible values of xy

33
53
83
35
55
85

Total cases = 9, because \(x\) can be filled in 3 ways and \(y\) can be filled in 3 ways

Here are all the possible values of xy

33
53
83
35
55
85
38
58
88

Hence required probability = \(\frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{6}{9} = \frac{2}{3}\)

 ­

­Crystal clear thank you! I was not comprehending that 53x,y12 was the six digit number... my mind split it into "53x" , "y12"... no wonder this problem was so confusing to me. Definitely attainable if you understand divisibility rules and some combinatorics.
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Re: Let s be the set of all 6-digit numbers of the form 54xy12, where each [#permalink]
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