Latest Elitmus Aptitude Question SOLUTION: how many six digit numbers can be formed using the digits 0 to 5,without repetition such that the number is divisible by the digit at its units place?

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(#M40010354) ELITMUS Numerical Ability question Keep an EYE

how many six digit numbers can be formed using the digits 0 to 5,without repetition such that the number is divisible by the digit at its units place?

Asked In Elitmus deepika (11 years ago)

Unsolved Read Solution (19)

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Ashiwani Kumar  (11 years ago)

the number ends with 1,2,3,4,5
so how many number can be made using given six digits whose last value is 1.
4*4*3*2*1 1(last digit is fix)=96
simillarly for 2,3,4,5 we will get total 96.
so total nuber that we can form is 96**5=480;
96*5=480;
in every case it is divisible by last unit place's num except 4;
in 4 some of the number that is not divisible are given bellow
the number which ends with 14,34,54
so for this
3*3*2*1 (14) is fixed
3*3*2*1 (34) is fixed
3*3*2*1 (54) is fixed
so total is 3*(3*3*2*1)=54

so final solution is:- 480-54=426

Is this solution Helpfull? Yes (68) | No (7)  

KRISHNA CHOUDHARY  (10 years ago)

when 0 is the last digit=0
when 1 is the last digit=4*4*3*2*1*1=96
when 2 is the last digit=4*4*3*2*1*1=96
when 3 is the last digit=4*4*3*2*1*1=96
when 4 is the last digit=(2,4)3*3*2*1*1*1=18 and(0,4)4*3*2*1*1*1=24=18+24=42
when 5 is the last digit=4*4*3*2*1*1=96
total=0+96+96+96+42+96=426 ANS

Is this solution Helpfull? Yes (47) | No (2)  

Garima  (11 years ago)

last digit can be 1,2,3,5,04,24 .
when last digit can be 1,2,3,5,
then no. of possible numbers = 4*5*4*3*2 = 480
when last digits are 04 and 24,
then no. of possible numbers = 1*2*4*3*2 = 48

total possible numbers = 480+48 = 528

Is this solution Helpfull? Yes (10) | No (30)  

Vaithiyanaathan  (8 years ago)

To begin with.
Step 1- Let 0 be the last digit. Obviously there are 0 possibilities for a number to be divisible by 0.

Step 2. Let 1 be the last digit. It has 96 possibilities. Totally if 1 is the last digit it has 120 possibilities. But 0 cannot be the first digit. So we have to subtract those possibilities.
Hence if 1 is last digit 120-24= 96 possibilities.

Step 3: If 2 is the last digit, it has 120 possibilities. But again 0 cannot be the first digit.
So 120- 24=96 possibilities.

Step 4: If 3 is the last digit, it again has 120 possibilities. But then again 0 cannot be the first digit. So again 120-24=96 possibilities.

Step 5: If 4 is the last digit, then either 0 or 2 must be the tens digit. If 0 is the tens digit , then it has 24 possibilities.
If 2 is the tens digit , it has 24 possibilities but 0 cannot be the first digit. So
24-6= 18 possibilities.
Step 6: If 5 is the last digit it has 120 possibilities. But again 0 cant be the first digit.
So 120-24=96
The final answer is 96+96+96+96+42= 426 possibilities.

Is this solution Helpfull? Yes (10) | No (1)  

Tarun Sharma  (8 years ago)

1. 0 can not be at end of any no.(divisiblity criteria will be failed)
2.number of combination using 1,2,3,5 at the end(becs each number ending with these digit will divisible by them)
all ending with 1 is divisible by 1
all ending with 2 is even no..and so divisible by 2
all ending with 3 has sum divisible by 3 so no is divisible by 3
all ending with 5 .so divisible by 5

i.e 4*4*3*2*1*1 =96, so 96*4=384
3.now for 4 we will check last 2 digits so can be 04 and 24
number of combination using 04 as last 2 digit .i.e 4*3*2*1=24
number of combination using 24 as last 2 digit i.e 3*3*2*1=18
so 384+24+18=426

Is this solution Helpfull? Yes (8) | No   

Abhishek Shandilya  (9 years ago)

solution: take 5 at unit place the _ _ _ _ _ 5 thn it is 4*4! and nw take 4 thn it have two cases _ _ _ _ 0 4 and _ _ _ _ 2 4 thn it is 4*3!+3*3! and again for 3 _ _ _ _ _ 3 thn it is 4*4! and nw for 2 it is _ _ _ _ _ 2 so 4*4! and similarly for 1 4*4! so we add oll (4*4!)+(4*3!+3*3!)+(4*4!)+(4*4!)+(4*4!)= 96+(24+18)+96+96+96= 426

Is this solution Helpfull? Yes (5) | No   

Ashutosh Anshu  (10 years ago)

The number ends with 0,1,2,3,4,5 .
so how many number can be made using given six digits whose last value is 1.
4*4*3*2*1 1(last digit is fix)=96
similarly for 2,3,5 we will get total 96.
so total number that we can form is 96*4=384;
in every case it is divisible by last unit place's number except 4;
in 4 some of the number that is divisible are given below
the number which ends with 04,24,44. But 44 can't be possible because repetition is not allowed.
so for this
3*3*2*1 (04) is fixed
3*3*2*1 (24) is fixed

so total is 2*(3*3*2*1)=36

so final solution is:- 384+36=420

Is this solution Helpfull? Yes (3) | No (10)  

vignesh  (11 years ago)

477 is the answer.
for 1 96 values can be formed
for 2 96 values can b formed.
for 3 also 96 values can be formed
for 5 96 values can b formed.
but for 4 93 values can be formed.(54,14,34)
there4 totlay 477 values can b formed.

Is this solution Helpfull? Yes (2) | No (22)  

suvadip rang  (10 years ago)

what about the unit place with the number 0?? it can not be divisible by 0..
numbers having 0 as unit digit= 5!
and also, numbers starting with 0 = 5!
and also, the number which ends with 14,34,54
so for this 4!* 3C1

so, total numbers= 6! - (4! * 3C1)- 5! - 5! = 408

Is this solution Helpfull? Yes (2) | No (7)  

Avik Saha  (11 years ago)

720 - (120+120+72) = 408

Is this solution Helpfull? Yes (1) | No (14)  

Md Asif Ashraf  (10 years ago)

Most of the solution has a common mistake while finding divisiblity by 4, and that is when we fix last two digit as |-|-|-|-|-|0|4| then there is a twist in the solution...
lets see the solution..
case1: if no ends with 1,2,3,5,6
we can arrange nos as: 4*4*3*2*1=96 (when last digit is any one of 1,2,3,5,6)
at the end we have total nos=5*96=480;
case2: if no ends with 4(always remember a no is divisible by 4 only when its last two digit is divisible by 4)
by given sequence here are three possiblities
ends with 04
ends with 24
ends with 64
subcase 1:ends with 04
no is: 4*3*2*1=24(becoz 0 is already fixed at second digitposition)
subcase 2:ends with 24
no is:3*3*2*1=18
subcase 3:ends with 64
no is:3*3*2*1=18
hence finally we have total possiblities=(480+24+18+18)=540

Is this solution Helpfull? Yes (1) | No (11)  

Sushant Shrikant Hiwale  (8 years ago)

all no. ending with 1 is divisible by 1
all no. ending with 2 is even no..and so divisible by 2
all no. ending with 3 has sum divisible by 3 so no is divisible by 3
all no. is divisible by 4 only when its last two digit is divisible by 4
all no. ending with 5 .so divisible by 5

Six digit number with the last digit 1:

4 * 4 * 3 * 2 * 1 * 1 =96
(out of 2,3,4,5) (0 + 3 remaining nos.) (3 rem.) (2 rem.) (1 rem.) (Fixed number)

Same for ending with 2 3 5
Now 4 * 96 = 384

ending with 4 only 04, 24

3*3*2*1 (04) is fixed
3*3*2*1 (24) is fixed

so total is 2*(3*3*2*1)=36

so final solution is:- 384+36=420

Is this solution Helpfull? Yes | No (3)  

sharwan maheshwari  (8 years ago)

i can,t understand case of when last digit is 4 please help

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nitish saxena  (7 years ago)

@SHARWAN MAHESHWARI: Check the divisibilty of 4... u better be understand..

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sandeep  (7 years ago)

Unit place digits-- 0,1,2,3,4,5
**0 digit not possible in unit place according to divisibility
Unit place digit 1-- 4*4*3*2*1*1=96
Unit place digit 2-- 4*4*3*2*1*1=96
Unit place digit 3-- 4*4*3*2*1*1=96
Unit place digit 5-- 4*4*3*2*1*1=96
unit place digit 4 --(according to divisibility rule --- last to digit check)--- 04, 24
04)--4*3*2*1*1*1(04 are fixed in last two digits so 1*1)=24
24)--3*3*2*1*1*1(24 are fixed in last two digits so 1*1)=18
So total number of ways=(96*4)+(24+18)=426

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sumit kumar  (7 years ago)

when 0 is the last digit=0
when 1 is the last digit=4*4*3*2*1*1=96
when 2 is the last digit=4*4*3*2*1*1=96
when 3 is the last digit=4*4*3*2*1*1=96
when 4 is the last digit=(2,4)3*3*2*1*1*1=18 and(0,4)4*3*2*1*1*1=24=18+24=42
when 5 is the last digit=4*4*3*2*1*1=96
total=0+96+96+96+42+96=426 ans

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Praveen  (7 years ago)

divisible by 0=0
divisible by 1=4*4*3*2*1=96
divisible by 2=4*4*3*2*1=96
divisible by 3=4*4*3*2*1=96
divisible by 4=(4*3*2*1)+(3*3*2*1)=24+18=42
divisible by 5=4*4*3*2*1=96

Adding all we get 426

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Arun dingh  (5 years ago)

#vignesh
Does 54'34 and 14 divisible by 4!!

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  (5 years ago)

when 0 is the last digit=0
when 1 is the last digit=4*4*3*2*1*1=96
when 2 is the last digit=4*4*3*2*1*1=96
when 3 is the last digit=4*4*3*2*1*1=96
when 4 is the last digit=4*4*3*2*1*1=96
when 5 is the last digit=4*4*3*2*1*1=96
total 96+96+96+96+96=480
but in case of unit digit of '4' there are some number which cannot be divisible by 4 like 14,34,54.
so, we will subtract them,
_ _ _ _14=3*3*2*1=18
_ _ _ _34=3*3*2*1=18
_ _ _ _54=3*3*2*1=18
so total is 18+18+18=54
total number is 480-54=426

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