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How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such
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21 Jul 2018, 04:58
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How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3digit number formed is a multiple of 3? (A) 6 (B) 24 (C) 27 (D) 28 (E) 64 New question !!!..
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Re: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such
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21 Jul 2018, 05:18
chetan2u wrote: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3digit 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 3digit 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 3digit numbers can be formed from digits 3, 6, 8 and 9 such
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21 Jul 2018, 05:32
chetan2u wrote: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3digit 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 3digit numbers a multiple of 3. 1. Considering the set {3,6,9}, We have \(3^3\) nos of 3digit 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 3digit number=27+1=28 Ans. (D)
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Re: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such
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23 Jul 2018, 07:22
souvonik2k wrote: chetan2u wrote: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3digit 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 3digit 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 3digit 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 3digit numbers can be formed from digits 3, 6, 8 and 9 such
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23 Jul 2018, 07:51
dave13 wrote: souvonik2k wrote: chetan2u wrote: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3digit 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 3digit 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 3digit 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/divisibilitymultiplesfactorstipsandhints174998.html?hilit=Divisibility%20by%209
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Re: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such
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23 Jul 2018, 08:54
PKN wrote: chetan2u wrote: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3digit 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 3digit numbers a multiple of 3. 1. Considering the set {3,6,9}, We have \(3^3\) nos of 3digit 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 3digit 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
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Re: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such
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23 Jul 2018, 09:07
dave13 wrote: PKN wrote: chetan2u wrote: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3digit 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 3digit numbers a multiple of 3. 1. Considering the set {3,6,9}, We have \(3^3\) nos of 3digit 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 3digit 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
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Re: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such
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23 Jul 2018, 09:08
dave13 wrote: PKN wrote: chetan2u wrote: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3digit 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 3digit numbers a multiple of 3. 1. Considering the set {3,6,9}, We have \(3^3\) nos of 3digit 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 3digit 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 3digit numbers can be formed from digits 3, 6, 8 and 9 such
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23 Jul 2018, 09:18
dave13 wrote: PKN wrote: chetan2u wrote: How many 3digit numbers can be formed from digits 3, 6, 8 and 9 such that the 3digit 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 3digit numbers a multiple of 3. 1. Considering the set {3,6,9}, We have \(3^3\) nos of 3digit 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 3digit 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 3digit 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 3digit number), hence no of permutations/arrangements=\(3^3\) 2. If repetition of digits is not allowed then the number of possible 3digits arrangements =nPn=3P3=3!
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