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# How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such

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Math Expert
Joined: 02 Aug 2009
Posts: 6565
How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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21 Jul 2018, 04:58
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Difficulty:

95% (hard)

Question Stats:

25% (01:54) correct 75% (01:28) wrong based on 84 sessions

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How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3-digit number formed is a multiple of 3?
(A) 6
(B) 24
(C) 27
(D) 28
(E) 64

New question !!!..

_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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21 Jul 2018, 05:18
2
chetan2u wrote:
How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3-digit number formed is a multiple of 3?
(A) 6
(B) 24
(C) 27
(D) 28
(E) 64

New question !!!..

Multiple of 3 means the sum of the digits of the 3-digit number must be divisible by 3.
We observe that if the no. contains 8, it cannot form a no. divisible by 3 except when all the digits are 8 i.e. 888.
(Check 3+6+8=17, 6+9+8=23 etc...)
Nos. that can be formed by digits 3,6,9 are 3x3x3=27, since each digit can be formed by these 3 digits to obtain a no. divisible by 3.
But we have to count the possibility of 888, so total nos. = 27+1=28.
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Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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21 Jul 2018, 05:32
2
chetan2u wrote:
How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3-digit number formed is a multiple of 3?
(A) 6
(B) 24
(C) 27
(D) 28
(E) 64

New question !!!..

Here we have two sets of numbers viz, {3,6,9} and {8}; those form 3-digit numbers a multiple of 3.

1. Considering the set {3,6,9}, We have $$3^3$$ nos of 3-digit numbers a multiple of 3. (Since it's not explicitly mentioned; We have to treat that repetition is allowed)
2. Considering the set {8}, We have only 1 three digit number(888) multiple of 3. (8+8+8=24 is a multiple of 3)

So, total no of 3-digit number=27+1=28

Ans. (D)
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PKN

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Joined: 09 Mar 2016
Posts: 773
Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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23 Jul 2018, 07:22
souvonik2k wrote:
chetan2u wrote:
How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3-digit number formed is a multiple of 3?
(A) 6
(B) 24
(C) 27
(D) 28
(E) 64

New question !!!..

Multiple of 3 means the sum of the digits of the 3-digit number must be divisible by 3.
We observe that if the no. contains 8, it cannot form a no. divisible by 3 except when all the digits are 8 i.e. 888.
(Check 3+6+8=17, 6+9+8=23 etc...)
Nos. that can be formed by digits 3,6,9 are 3x3x3=27, since each digit can be formed by these 3 digits to obtain a no. divisible by 3.
But we have to count the possibility of 888, so total nos. = 27+1=28.

souvonik2k Hi there

Why do you say "Multiple of 3 means the sum of the digits of the 3-digit number must be divisible by 3.

Where is it mentoned 'the sum of the digits" it could be PRODUCT OF THREE DIGITS as well

thanks,

D.
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Math Expert
Joined: 02 Aug 2009
Posts: 6565
Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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23 Jul 2018, 07:51
1
dave13 wrote:
souvonik2k wrote:
chetan2u wrote:
How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3-digit number formed is a multiple of 3?
(A) 6
(B) 24
(C) 27
(D) 28
(E) 64

New question !!!..

Multiple of 3 means the sum of the digits of the 3-digit number must be divisible by 3.
We observe that if the no. contains 8, it cannot form a no. divisible by 3 except when all the digits are 8 i.e. 888.
(Check 3+6+8=17, 6+9+8=23 etc...)
Nos. that can be formed by digits 3,6,9 are 3x3x3=27, since each digit can be formed by these 3 digits to obtain a no. divisible by 3.
But we have to count the possibility of 888, so total nos. = 27+1=28.

souvonik2k Hi there

Why do you say "Multiple of 3 means the sum of the digits of the 3-digit number must be divisible by 3.

Where is it mentoned 'the sum of the digits" it could be PRODUCT OF THREE DIGITS as well

thanks,

D.

Hi..

There are certain divisibility rules ..
For 3 and 9, it is that if sum of the digit is div by 3 or 9, the numy is divisible by 3 or 9..
Example..
363...3+6+3=12 div by 3, do 363 is div by 3 but not by 9 as 12 is not div by 9...
999999999999 will be div by both 3 and 9

Some more rules can be seen in the following post.
https://gmatclub.com/forum/divisibility-multiples-factors-tips-and-hints-174998.html?hilit=Divisibility%20by%209
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Director
Joined: 09 Mar 2016
Posts: 773
Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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23 Jul 2018, 08:54
PKN wrote:
chetan2u wrote:
How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3-digit number formed is a multiple of 3?
(A) 6
(B) 24
(C) 27
(D) 28
(E) 64

New question !!!..

Here we have two sets of numbers viz, {3,6,9} and {8}; those form 3-digit numbers a multiple of 3.

1. Considering the set {3,6,9}, We have $$3^3$$ nos of 3-digit numbers a multiple of 3. (Since it's not explicitly mentioned; We have to treat that repetition is allowed)
2. Considering the set {8}, We have only 1 three digit number(888) multiple of 3. (8+8+8=24 is a multiple of 3)

So, total no of 3-digit number=27+1=28

Ans. (D)

hey there PKN if repetition is allowed (and you wrote 888) then how about number 333 3+3+3 =9 is a multtiple of 3

and how about 9+9+9 = 27 is multiple of 3 so whats wrong then with my reasoning now
_________________

In English I speak with a dictionary, and with people I am shy.

Math Expert
Joined: 02 Aug 2009
Posts: 6565
Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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23 Jul 2018, 09:07
1
dave13 wrote:
PKN wrote:
chetan2u wrote:
How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3-digit number formed is a multiple of 3?
(A) 6
(B) 24
(C) 27
(D) 28
(E) 64

New question !!!..

Here we have two sets of numbers viz, {3,6,9} and {8}; those form 3-digit numbers a multiple of 3.

1. Considering the set {3,6,9}, We have $$3^3$$ nos of 3-digit numbers a multiple of 3. (Since it's not explicitly mentioned; We have to treat that repetition is allowed)
2. Considering the set {8}, We have only 1 three digit number(888) multiple of 3. (8+8+8=24 is a multiple of 3)

So, total no of 3-digit number=27+1=28

Ans. (D)

hey there PKN if repetition is allowed (and you wrote 888) then how about number 333 3+3+3 =9 is a multtiple of 3

and how about 9+9+9 = 27 is multiple of 3 so whats wrong then with my reasoning now

Dave..
When you write 3*3*3, it includes all possibilities..
Each 3 in 3*3*3 means that any of the three digits 3,6,9 can be used..
So it includes 333,336,339,369,396,366 and so on
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Status: Preparing for GMAT
Joined: 25 Nov 2015
Posts: 758
Location: India
GPA: 3.64
Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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23 Jul 2018, 09:08
1
dave13 wrote:
PKN wrote:
chetan2u wrote:
How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3-digit number formed is a multiple of 3?
(A) 6
(B) 24
(C) 27
(D) 28
(E) 64

New question !!!..

Here we have two sets of numbers viz, {3,6,9} and {8}; those form 3-digit numbers a multiple of 3.

1. Considering the set {3,6,9}, We have $$3^3$$ nos of 3-digit numbers a multiple of 3. (Since it's not explicitly mentioned; We have to treat that repetition is allowed)
2. Considering the set {8}, We have only 1 three digit number(888) multiple of 3. (8+8+8=24 is a multiple of 3)

So, total no of 3-digit number=27+1=28

Ans. (D)

hey there PKN if repetition is allowed (and you wrote 888) then how about number 333 3+3+3 =9 is a multtiple of 3

and how about 9+9+9 = 27 is multiple of 3 so whats wrong then with my reasoning now

Hi
333, 999 and 666 are already covered in 3^3=27, but 888 is not covered. So it is added and answer is 27+1=28.
Hope it is clear.
_________________

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When the going gets tough, the tough gets going...

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Status: Learning stage
Joined: 01 Oct 2017
Posts: 546
WE: Supply Chain Management (Energy and Utilities)
Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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23 Jul 2018, 09:18
1
dave13 wrote:
PKN wrote:
chetan2u wrote:
How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3-digit number formed is a multiple of 3?
(A) 6
(B) 24
(C) 27
(D) 28
(E) 64

New question !!!..

Here we have two sets of numbers viz, {3,6,9} and {8}; those form 3-digit numbers a multiple of 3.

1. Considering the set {3,6,9}, We have $$3^3$$ nos of 3-digit numbers a multiple of 3. (Since it's not explicitly mentioned; We have to treat that repetition is allowed)
2. Considering the set {8}, We have only 1 three digit number(888) multiple of 3. (8+8+8=24 is a multiple of 3)

So, total no of 3-digit number=27+1=28

Ans. (D)

hey there PKN if repetition is allowed (and you wrote 888) then how about number 333 3+3+3 =9 is a multiple of 3

and how about 9+9+9 = 27 is multiple of 3 so whats wrong then with my reasoning now

Hi dave13,
333, 999,666 these are already included in 3^3 nos of 3-digit numbers ,a multiple of 3.
Had it been a case of "repetition is not allowed" then we had to exclude 333,666,999,888,336,363,663 etc.....so on...

Thanking You.

P.S:-
1. The number of permutations of 'n' different things taken 'r' at a time , when each is allowed to repeat any number of times in each arrangement is $$n^r$$.
Here 3 different digits{3,6,9} are to be taken 3 at a time(since we have to form a 3-digit number), hence no of permutations/arrangements=$$3^3$$

2. If repetition of digits is not allowed then the number of possible 3-digits arrangements =nPn=3P3=3!
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such &nbs [#permalink] 23 Jul 2018, 09:18
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