A seven-digit number comprises of only 2's and 3's. How many of these are multiples `of 12?
(A)1
(B)11
(C)21
(D)44

• (B) 11
  for a number to be divisible by 12, it should be divisible by 3 and 4 both.
  last two digit will be 32 only.
  -----32
  sum of digits should be multiple of 3, last two digit sum is 5 so first 5 digit sum can be 10,13 only.
  so, 1st 5 digit can be 22222 or 22333 .
  
  numbers are 2222232 , 2233332 & its permutation.
  total no.s = 1 + 5!/(3!*2!) = 11
  
• 10 years agoHelpfull: Yes(34) No(2)
• for the number to be divisible by 12 it has to be divisible by 4 and 3 both
  1. to be divisible by 4 last two digits have to be 32
  -----32
  2. to be divisible by 3
  case 1 : all digits be 2 i.e 2222232
  case 2 : three 2's and two 3's be arranged in 5!/(2!*3!) = 10
  case 3 : two 3's and three 3's be arranged in 5!/(2!*3!) =10
  therefore total number of ways =1+10+10=21
• 10 years agoHelpfull: Yes(9) No(16)
• 11 solutions are available. 3332232, 3323232, 3233232, 2333232, 3322332,3223332, 2233332, 3232332, 232332, 2332332, 2222232
• 10 years agoHelpfull: Yes(6) No(0)
• @ Sonu 3332232 is also possible .There will be many more.
• 10 years agoHelpfull: Yes(3) No(1)
• again calculation mistake its 10+1=11
• 10 years agoHelpfull: Yes(3) No(0)
• Answer is 11 only...
  for divisibility by 4...last two digit has to be 32...
  now seven digits look like -----32
  case1: When all first five digits are 2
  case 2:when we have two 2s and three 3's coz it makes sum 18 so divisible by 3
  this will be 5!/(2!*3!)=10
  therfore 10+1=11
• 10 years agoHelpfull: Yes(3) No(1)
• only 1 such number is possible...2222232 because factors of 12=3*4 last two digits must be divisible by 4 so it happens only if last two digits are 32 and for 3 sum of all digit is divisible by 3 so 2spots are already filled rest are 5 ,sum of 3+2=5 i.e sum of first five digit is either 10 or 13 so sum become 15 or 18 ,second case is not possible , first case is possible if first five digits are 2's .....
• 10 years agoHelpfull: Yes(2) No(2)
• sorry case 3 will not be there
  ans is 1+10=11 only
• 10 years agoHelpfull: Yes(2) No(0)
• 11 is ryte ans
  coz only 2 cases r possile
  1-2222232
  2-2233332
  n in case 2 we arrange the no by its permutation
  5!/3!*2!=10
  then total caes =10+1=11
• 10 years agoHelpfull: Yes(1) No(0)
• 21 is the ans , first num is 2222232, 3332232 is also possible as last two digits are fixed so first five digits can arrange them in 5!/3!*2!=20 , 20+1=21 ,
• 10 years agoHelpfull: Yes(0) No(6)
• 22222 = 1
  two(2)and three(3)=10
  three(2) and three 3=10
  
  total 21 answer
• 10 years agoHelpfull: Yes(0) No(3)
• I think 21 is correct answer...
• 10 years agoHelpfull: Yes(0) No(1)
• ans. is 21 as there are 2 cases case 1. when no. ends with 21 nd case 2. when no. ends with 22 now for its divisibility by 3 there are another 2 cases when sum is either 15 or 18
• 10 years agoHelpfull: Yes(0) No(1)
• solution are the form of
  -----22
  then 1st five position filled by
  sum of first five =13
  then the only combination =2 3's and 3 2's
  the permuatution are
  5!/3!2!=10
  and one combination is 2222232
  so total=1+10=11 ans
• 10 years agoHelpfull: Yes(0) No(0)
• solution are the form of
  -----22
  then 1st five position filled by
  sum of first five =14
  then the only combination =2 3's and 3 2's
  the permuatution are
  5!/3!2!=10
  and one combination is 2222232
  so total=1+10=11 ans
• 10 years agoHelpfull: Yes(0) No(0)
• Answer is 11
  Number should be a multiple of 3 and 4.
  So, the sum of the digits should be a multiple of 3.
  
  We can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3 (The number of 2's should be a multiple of 3).
  
  For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.
  
  (i)- All seven 3's - No possibility
  
  (ii)- Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some order. No of possibilities =
  5!
  3!×2!
  =10
  
  (iii)- Six 2's and one 3 - The first 5 digits should all be 2's.
  So, there is - only one number - 2222232
  
  So we get the total of seven-digit number comprises of only 2's and 3's which are multiples of 12 are:
  = (i) + (ii) + (iii)
  = 0+10+1=11
• 8 years agoHelpfull: Yes(0) No(0)