How many three digit number formed by using 2,3,4,5,6,7 once such that number is divisible by 15.
a) 8
b) 16
c)12
d)20

• If a no. is divisible by 15, so it is also divisible by 3 and 5
  for a no. to be divisible by 5 the unit digit must be 0 & 5 but 0 is not given so unit digit is 5.
  now use the divisibility test of 3 hence we get 8 different no.s which are
  345 435
  375 735
  465 645
  675 765
• 7 years agoHelpfull: Yes(36) No(8)
• The rule of a number divisible by 15 is the number is end with 5 and the sum of digits is divisible by 3 also. Here 5 is fixed at the last position and remaining 5 digits can fill 10th and 100th place in 8 different ways such that
  (3,4)
  (3,7)
  (4,3)
  (4,6)
  (6,4)
  (6,7)
  (7,3)
  (7,6)
  So the ans is 8.
• 7 years agoHelpfull: Yes(14) No(2)
• 4*2*1=8
  at one's place only 5 will come
  and at ten's place 4 and 7 can be placed,
  and at 100th place rest of the 4 digits can come...
  so the answer is 8
• 7 years agoHelpfull: Yes(11) No(10)
• unit digit must be 5 i.e _ _ 5 (only one number in unit place)
  Now, we have 5 more number that can be in 10th place that are 2,3,4,6,7 (5 will not because 5 has already used )
  so we have _ (any one from 2,3,4,6,7) 5
  now we have only 4 numbers to be used on 100th place
  so we 4*5*1=20
  
• 7 years agoHelpfull: Yes(7) No(15)
• 1- no divisible by 15 should be divisiblled by both 5 and 3 .
  2-divisibility conditions for 5 is the unite place should be 5.
  3-now we have to check how many numbers are there which will be divisible by 3 with unite place 5 of that no .
  4-for check the no is divisible by 3 , the summation of digits should be divisible by 3 .
  5- manually you can check
  these option would be (4 3 5) (3 4 5)(5 2 5 )(2 5 2)(7 3 5)(3 7 5)(6 4 6)(4 6 4)(7 6 5)(6 7 5)(2 2 5)(555)
  so there wil be 12 options .
• 7 years agoHelpfull: Yes(4) No(20)
• THEN IT IS 8 BROTHER .#abhijit jain
• 7 years agoHelpfull: Yes(4) No(1)
• 235 345 465 675 725
  245 365 475 625 735
  265 375 425 635 745
  275 325 435 645 765 ans :20
• 7 years agoHelpfull: Yes(4) No(52)
• 8 is the correct answer. we cant use a digit more than once. so 8 will be the only valid answer here.
• 7 years agoHelpfull: Yes(3) No(0)
• sibabrat , in the question it is clearly mentioned that repetition of digits is not allowed.
  
• 7 years agoHelpfull: Yes(2) No(0)
• There are 6 number take it unit ,tens and hindered . We know that to divide by 15 unit should be 5 and it is 1way and remaining 5 will in two ways 5c2 than 20 is ans
• 7 years agoHelpfull: Yes(2) No(3)
• if no. is divisible by 15 at once than last two digit must be
  (15,30,45,60,75,90.......135)
  but here only two choice available either 45 or 75.
  so first case
  when last two digits 45 then
  no.of three digits =6
  and when last two digits =75
  then no. of three digits=6
  add both=6+6
  =12
  ans=12
• 7 years agoHelpfull: Yes(2) No(5)
• 8
  the no. formed which should be divisible by 15 means it must be divisible by 3 and 5
  than unit place should be 5 (for divisibility of 5)
  for divisibility of 3
  345
  435
  375
  735
  465
  645
  675
  765
• 7 years agoHelpfull: Yes(2) No(0)
• MONIKA divisibility criteria of 15 is not only unit place should be 5. Unite place is 5 then it is divisible by 5 only .
• 7 years agoHelpfull: Yes(0) No(0)
• ans: 8 because there is no repition , for divisible of unit place should be 5 and for divisible of 3 sum of all digit should b divisible by 3
• 7 years agoHelpfull: Yes(0) No(0)
• opt. c 12
  divisible no. are
  135,315,165,615,
  345,435,375,735,
  465,645,675,765
• 7 years agoHelpfull: Yes(0) No(8)
• 1- no divisible by 15 should be divisiblled by both 5 and 3 .
  2-divisibility conditions for 5 is the unite place should be 5.
  3-now we have to check how many numbers are there which will be divisible by 3 with unite place 5 of that no .
  4-for check the no is divisible by 3 , the summation of digits should be divisible by 3 .
  5- manually you can check
  these option would be (4 3 5) (3 4 5)(5 2 5 )(2 5 2)(7 3 5)(3 7 5)(6 4 6)(4 6 4)(7 6 5)(6 7 5)(2 2 5)(555)
  so there wil be 12 options .
• 7 years agoHelpfull: Yes(0) No(5)
• 3p5=120
  120/15=8
  
• 7 years agoHelpfull: Yes(0) No(1)
• which one is correct ans ? 8 or 20 ?
• 7 years agoHelpfull: Yes(0) No(0)
• sorry for the last answer i.e 5! i.e 120 /15 =8
• 7 years agoHelpfull: Yes(0) No(0)
• 5 gets fixed in unit place now we require a sum of 4 ,7,10,13 using other digits to make it divisible by 3 (4+5=9 ,7+5=12.....) which satisfy divisibility by 3.
  Now possible ways for
  4 =(2,2)
  7=(5,2) (2,5) ,(4,3) ,(3,4)
  10=(5,5) ,(6,4) ,(4,6) , (7,3) ,(3,7)
  13=(7,6) ,(6,7)
  Total =12
• 7 years agoHelpfull: Yes(0) No(1)
• in order to make the number divisible by 15..the last digit should be 5 so that it is divisible by 5 ..and the summation of the numbers is (2+3+4+6+7+5=27)i.e divisible by 3 hence the last digit is kept fixed to 5 n the 1st 2 digits would be made with 2,3,4,6,7..so ans is5C2=20
  
• 7 years agoHelpfull: Yes(0) No(0)
• b) 8 is right answer
  three digit number divisible by 15 so number must be divisible by 3 and 5. so 5 is fixed on unit digit . and we have to consider tenth and thousand place respectively so that number could be divisible by 3.
  so pair is (3,4), (4,3), (3,7), (7,3), (4,6), (6,4), (6,7), (7,6).
• 7 years agoHelpfull: Yes(0) No(0)
• a number is divisible by 15 if last digit is either 5 or 0...so last digit can be filled in one way ie 5.other 2 digits can be formed from 5 digits in 5p2 ways.ie 20.
• 7 years agoHelpfull: Yes(0) No(0)
• Answer is 8,
  If number is divisible by 15 then it should be divisible by both 3 and 5,
  So, for a number to be divisible by 5, it's last digit should be either 5 or 0. Since 0 is not given, unit digit is 5,
  now for a number to be divisible by 3, the sum of digits of number should be divisible by 3.
  So, we can avoid numbers that include 2, so there are only 12 different combinations. Out of which we can eliminate 4 combinations (3,6,5), (4,7,5), (6,3,5), (7,4,5) as their digit summation is not multiple of 3. So, there are only 8 different combinations.
• 7 years agoHelpfull: Yes(0) No(0)
• Any number which is divisible by 15 .
  The number must be divisible by both 5 and 3
  First we written numbers which are divisible by 5 ( because every number which is divisible by 5 , last must be zero or 5 )
  5*4*1=20(total 20 numbers - - 5)
  Now check in 20 numbers which is divisible by 3
  Rule of sum of all digit of a number is n.
  n%3=0
  .......
  ........
  check all numbers and after checking answer is ::8
• 7 years agoHelpfull: Yes(0) No(0)
• If a number is divisible by 15 it must be divisible by both 3 and 5; bcz 3x5=15;
  So, a 3 digit number divisible by 5 must have 5 or 0 at unit place : __ __ 5 (here 0 is unavailable)
  a 3 digit number divisible by 3 must have sum of digits, divisible by 3.
  The only options are : 3 4 , 3 7, 4 6, 6 7 ; i,e, total 4 options
  all have two combinations
  So 4x2 = 8
  Ans : a) 8
• 7 years agoHelpfull: Yes(0) No(0)
• In a three digit number, the last digit is always 5 and 0 but 0 is not given so ignore 0. Now the remaining 100ths place will be formed by 5 ways and 10ths place will be formed by 4 ways.
  Finally, the result is 5*4*1 i.e., 20 ANSWER.
• 6 years agoHelpfull: Yes(0) No(1)