How many five digit multiples of 3 are formed using 0, 1, 2, 3, 4, 6 and 7 without repetition?


1) 312
2) 3125
3) 120
4) 216

• 0,1,2,3,4,6,7 - seven digits are given. we need only 5 digits.
  if a number to be divisible by 3.the sum of their digits must be divisible by 3.
  0+1+2+3+4+6+7 = 23
  we have to leave two digits, we have to select 5 out 0f 7 ,7c5 = 35 ways.
  minimum pair {0,1},sum = 1
  maximum pair {6,7},sum=13
  sum varies from 1 to 13.
  i.e)23-1 to 23-13 =22 to 10
  in that sum of 21,18,15,12 are divisible
  pairs that produces sum of 21 is {0,2}
  {1,3,4,6,7} 5!=120
  pairs that produces sum of 18 are {1,4} & {2,3}
  {0,2,3,6,7} 5! - 4!(since the first digit should not be zero)=96
  {0,1,4,6,7} similarly 96
  pairs that produces sum of 15 are {1,7},{2,6}
  {0,2,3,4,6} 96
  {0,1,3,4,7} 96
  pairs that produces sum of 12 are {4,7}
  {0,1,2,3,6} 96
  
  Total no of numbers formed=120+(96*5)=600
• 10 years agoHelpfull: Yes(23) No(20)
• For 3 multiple the sum should be divisible by 3
  So {0,1,2,3,4,6,7} produces a set {1,3,4,6,7} which sum is 21 divisible by 3
  Set {1,3,4,6,7} has 5 combination then 5!= 120 answer is option c)120
• 10 years agoHelpfull: Yes(17) No(9)
• every one's approach is different and you all got different set of answers but, i didn't get the solution from these answers.
• 10 years agoHelpfull: Yes(12) No(2)
• 4×4×3×2×2=172
• 10 years agoHelpfull: Yes(5) No(9)
• if zero is left out 5!=120 and 3 is left out
  then it will be 4*4!=96
  total 216
  ans (4)
• 10 years agoHelpfull: Yes(3) No(10)
• it is 5 digit number so 1st digit is always start with 1,2,3,4,5... then if a number is divisible by 3 sum of their digits is always divisible by 3...
  the following chances are:
  1){0,1,2,3.6}=4*4*3*2*1=96
  2){1,3,4,6,7}=5*4*3*2*1=120
  3){0,3,5,6,7}=4*4*3*2*1=96
  4){0,2,3,6,7}=4*4*3*2*1=96
  5){0,1,3,4,7}=4*4*3*2*1=96
  6)(0,2,4,5,7}=4*4*3*2*1=96
  7){1,2,5,6,7}=5*4*3*2*1=120
  
  the total chances are=720
  therefore 720 five digits are formed using 0, 1, 2, 3, 4, 6 and 7 without repetition divisible by 3...
• 10 years agoHelpfull: Yes(3) No(6)
• if sum of 5 digits number is divisible by 3 then such number is also multiple of 3 .except o we can put any no on 10000's position .
  let us consider 10000's place will be fill by one and other 4 position will fill by other remaining digits,remaining digits also rearranged between each other by 5! ways.
  similarly we can put the all remaining 5 digits in 10000's place .
  so total no of such combination is 6*5!=600
• 9 years agoHelpfull: Yes(2) No(0)
• We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.
  
  We have six digits: 0,1,2,3,4,5. Their sum=15.
  
  For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.
  
  Second step:
  
  We have two set of numbers:
  {1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets:
  
  {1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.
  
  {0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=96
  
  120+96=216
• 9 years agoHelpfull: Yes(2) No(1)
• 0,1,2,3,4,6,7 -seven digits
  we have fill in five palaces
  first place- 5
  second place-5
  third place-4
  fourth place-3
  fifth place-2
  Therefore 5*5*4*3*2=600
• 9 years agoHelpfull: Yes(2) No(1)
• if 0 is used
  4*4*3*2*2=192
  if 1 is used 5!=120
  therefore total ways 312
• 10 years agoHelpfull: Yes(1) No(3)
• 312MULTIPLES
  
• 10 years agoHelpfull: Yes(1) No(1)
• 0+1+2+3+4+6+7=23
  the set of5 digits that could be possible is: (1,3,4,6,7) and (2,3,4,6,0)
  (1,3,4,6,7) can be arranged in 5! ways= 120
  (2,3,4,6,0) can be arranged in 4x4x3x2x1 ways= 96
  so no. of digits are= 120+96=216
  option d is correct.
• 9 years agoHelpfull: Yes(1) No(0)
• Correct me if i m wrong....
  Given set,{0,1,2,3,4,6,7}
  first set will be {0,1,2,3,6} - ways 96 (4*4*3*2*1)
  second set will be {1,3,4,6,7} - ways 120 (5!)
  third set will be {1,3,4,7,0}- ways 96 similar as above.
  Total ways= 96+96+120
  =>312
• 7 years agoHelpfull: Yes(1) No(0)
• for sum = 21 (i.e divisible by 3 no. are : 1,3,4,6,7,)
  for sum = 18 (no are (0,2,3,6,7) or(1,4,6,7,0))
  
  (5*4*3*2) + 2(4*4*3*2) = 312
• 6 years agoHelpfull: Yes(1) No(0)
• i agree with you poojitha all the answers are wrong
• 10 years agoHelpfull: Yes(0) No(2)
• irst step:
  We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.
  
  We have six digits: 0,1,2,3,4,5. Their sum=15.
  
  For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1, 2, 3, 4, 5} and 15-3={0, 1, 2, 4, 5}. Meaning that no other 5 from given six will total the number divisible by 3.
  
  Second step:
  
  We have two set of numbers:
  {1, 2, 3, 4, 5} and {0, 1, 2, 4, 5}. How many 5 digit numbers can be formed using this two sets:
  
  {1, 2, 3, 4, 5} --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.
  
  {0, 1, 2, 4, 5} --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=96
  
  120+96=216
  
  Answer: E.
• 9 years agoHelpfull: Yes(0) No(0)
• the answer has to be by method as used by shanmuga
• 7 years agoHelpfull: Yes(0) No(0)
• Required numbers are = 5! + 5! - 4! = 216
• 6 years agoHelpfull: Yes(0) No(0)