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

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

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New post 21 Jul 2018, 03:58
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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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New post 21 Jul 2018, 04:18
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.
Answer D.
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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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New post 21 Jul 2018, 04:32
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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Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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New post 23 Jul 2018, 06:22
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.
Answer D.



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 :?

please explain :)

thanks,

D. :-)
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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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New post 23 Jul 2018, 06:51
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.
Answer D.



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 :?

please explain :)

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


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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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New post 23 Jul 2018, 07:54
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 :-)
Math Expert
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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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New post 23 Jul 2018, 08:07
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


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User avatar
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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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New post 23 Jul 2018, 08:08
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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Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such  [#permalink]

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New post 23 Jul 2018, 08:18
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

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Re: How many 3-digit numbers can be formed from digits 3, 6, 8 and 9 such &nbs [#permalink] 23 Jul 2018, 08:18
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