How many 6 digit numbers can be formed using digits 1 to 6 without repetition such that the number is divisible by the digit at its units place?

We have to form 6 digit numbers using the digits 3, 4, 5, 6, 7, 8 without repetition.
Total number of ways of arranging 6 digits in six places = 6P6 = 6!= 6 × 5 × 4 × 3 × 2 × 1 = 720 waysHere, the number is divisible by 5. So it will have the digit 5 in the unit’s place. Hence, the unit’s place can be filled in 1 way.The other five places can be filled in by the remaining 5 digits

(Since repetition is not allowed) in 5P5 = 5! Ways.

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Elitmus Exam Numerical Ability Permutation and Combination

• Total possible nos without condition of divisibility = 6! = 720 only some nos with last digit 4 may not be divisible which are ending with 14,34,54 no of such nos are 3*4*3*2 = 72

  so six digit number can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place = 720-72= 648

• 10 years agoHelpfull: Yes(73) No(16)
• Ans. 648 As each number will contain all the six digits and the sum of digits is = 1+2+3+4+5+6 = 21 which is divisible by 3. So each number is divisible by 3. The numbers ending with digit 1 will be divisible 1. The numbers ending with digit 2 will be divisible 2. The numbers ending with digit 3 are divisible 3. The numbers ending with digit 5 will be divisible 5. The numbers ending with digit 6 will be divisible 6. Except the numbers ending with last two digits as 14, 34 and 54 all other numbers ending with 4 are divisible by 4. The no. of numbers ending with last two digits 14,34 and 54 are = 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4) so six digit number that can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place

  = 720-72= 648

• 10 years agoHelpfull: Yes(53) No(5)
• If units digit has 1 then possible numbers = 120(5!)---------->1 _ _ _ _ _ 1 (remaining 5 digits can be filled in 5! ways if units digit has 2 then possible numbers = 120-------------->2 if units digit has 3 then possible numbers = 120-------------->3 as the sum of all digits is 21 the number is divisible by 3 if units digit has 4 then possible numbers = 48 (4!*2) as the last two digits must 24,64 remaining 4 digits can be arranged in 4!*2 if units digit has 5 then possible numbers = 120 if units digit has 6 then possible numbers = 120

  total 120+120+120+48+120+120=648

• 8 years agoHelpfull: Yes(46) No(2)
• We will go through this step by step. Divisible by 1 and 1 at the unit place: _ _ _ _ _ 1 This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers. Divisible by 2 and 2 at the unit place: _ _ _ _ _ 2 This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers. Divisible by 3 and 3 at the unit place: _ _ _ _ _ 3 Since, any number will have all the digits from 1 to 6 and the sum 1 + 2 + 3 + 4 + 5 + 6 = 21 is divisible by 3 This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers. Divisible by 4 and 4 at the unit place: _ _ _ _ _ 4 Here, there are two cases _ _ _ _ 2 4 This gives 4 x 3 x 2 x 1 = 24 numbers. _ _ _ _ 6 4 This gives 4 x 3 x 2 x 1 = 24 numbers. This gives us total of 24 + 24 = 48 numbers. Divisible by 5 and 5 at the unit place: _ _ _ _ _ 5 This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers. Divisible by 6 and 6 at the unit place: _ _ _ _ _ 6 As all 6 digit numbers formed with 1 to 6 digits(without repetition) are divisible by 3 and numbers with 6 at the unit place are even. This gives us total of 5 x 4 x 3 x 2 x 1 = 120 numbers. None of these cases will have numbers overlapping with each other. So, Total numbers = 120 + 120 + 120 + 48 + 120 + 120 = 648
• 7 years agoHelpfull: Yes(31) No(0)
• I have gone with the cases putting 1 2 3 4 5 6 as unit digits
  and got the answer as 648(120+120+120+48+120+120)
• 10 years agoHelpfull: Yes(25) No(1)
• thanks arjun bhai....great thinking power
• 9 years agoHelpfull: Yes(14) No(0)
• How you are calculating for divisible by 4.
  I am not getting 3*4*3*2 = 72.from where 3 *4*3*2 is coming.
• 9 years agoHelpfull: Yes(10) No(2)
• 6!-4!*3C1=720-72=648
• 9 years agoHelpfull: Yes(7) No(1)
• ans:2468 1 as units place=720 2 as units place=360 3 as units place=720 4 as units place=168 5 as units place=120 6 as units place=360

  total six digit numbers=720+360+720+168+120+360=2468

• 10 years agoHelpfull: Yes(6) No(66)
• oops i missed something... If digit place has 1 then possible numbers = 120(5!) if 2 then possible numbers = 120 if 3 then possible numbers =120 if 4 then possible numbers = 48 (4!*2) if 5 then possible numbers = 120 if 6 then possible numbers = 120

  so total no=648 now it's right!

• 8 years agoHelpfull: Yes(6) No(1)
• I THINK 648 NUMBERS ARE POSSIBLE.....
• 10 years agoHelpfull: Yes(5) No(17)
• total possible no =6! ie 720 720/6=120 will b divisible each for 1,2,3,5,6 for 4 divisible its 2/5*120=48( as 14,24,34,44,54 only out of these only 2 numbers are divisible by 4)

  so total number =120+120+120+120+120+48=648

• 8 years agoHelpfull: Yes(4) No(0)
• 3*4*3*2*1 actually means 3*4! 3 for 14,34,54 which are not divisible by 4. 4! are the ways. So Ans is 720-3*4!=648
• 8 years agoHelpfull: Yes(3) No(0)
• If digit place has 1 then possible numbers = 120(5!) if 2 then possible numbers = 120 if 3 then possible numbers =120 if 4 then possible numbers = 48 (4!*2) if 5 then possible numbers = 120 if 6 then possible numbers = 20 so total no=648
• 8 years agoHelpfull: Yes(2) No(1)
• when unit place is 1 then 5*4*3*2*1=120 when unit place is 2 then 5*4*3*2*1=120 """""""""""""""""' 3 then 120 """""""""""""""""" 4 then 24 """""""""""""""""" 5 then 120 """""""""""""""""" 6 then 120

  total is 5*120+24=624

• 8 years agoHelpfull: Yes(2) No(1)
• when unit place is 1 then 5*4*3*2*1=120 when unit place is 2 then 5*4*3*2*1=120 """""""""""""""""' 3 then 120 """""""""""""""""" 4 then 24*2 bcuz 24 and 64 is divisible by 4 """""""""""""""""" 5 then 120 """""""""""""""""" 6 then 120

  total is 5*120+24=648

• 8 years agoHelpfull: Yes(2) No(0)
• Given number 1,2,3,4,5,6 when unit digit 1 then number ways=5*4*3*2*1=120 number way 1,2,3,5,6=120*5=600 if unit digit 4 then tens place 2,6=4*3*2=24*2=48

  total number of ways=600+48=648

• 6 years agoHelpfull: Yes(2) No(0)
• it should be 528 because the number that ends with 1,2,3,5 are divisible with 1,2,3,5 respectively but the numbers that ends with 4 and 6 has to have second last digit be 2, so with this we have 48 more numbers which gives total of 528...
• 8 years agoHelpfull: Yes(1) No(3)
• u all guys are including 6 bt ony 36 is divisible by 6

  wat abt 16,26,46,56?????

• 8 years agoHelpfull: Yes(1) No(1)
• digits are divisible by One are=1*5*4*3*2*1=120 (1, , , , ,(2 to 6) ) digits are divisible by Two are=1*4*3*2*1*2=48 (2, , , , ,(4 or 6) ) digits are divisible by Three are=1*5*4*3*2*1=120 (3, , , , ,(1 to 6 expect 3) ) digits are divisible by four are=1*4*3*2*2=48 (4, , , ,(last two digit may be 12 or 16 or 32 or 36 or 52 or 56) ) digits are divisible by Five are=0, because unit digit is 5 and in last should be 5 or 0, we can't repeat digits are divisible by Six are= divisible by 2 or 3 = 48+120=168

  120+48+120+48+0+168=528

• 7 years agoHelpfull: Yes(1) No(5)
• if the question was as like - "six digit no. can be formed using the digit 0 to 5,without repetition.......... .unit's place" ...then what will be answer a.420 b.426 3.432 d.none ???
• 8 years agoHelpfull: Yes(0) No(0)
• I have gone with the cases putting 1 2 3 4 5 6 as unit digits and got the answer as 648(120+120+120+48+120+120)

  Read more at http://www.m4maths.com/frequently-asked-placement-questions.php?SOURCE=elitmus&TOPIC=Numerical%20Ability&SUB_TOPIC=Permutation%20and%20Combination#vFws7SICk58fhqWe.99

• 7 years agoHelpfull: Yes(0) No(0)

How many 6 digit even numbers can be formed using the digits 1 to 6?

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How many 6 digit numbers can be formed by first 6 whole numbers with repetition which are divisible by 2?