M4maths Previous Puzzles

28 Dec 2016 Permutation

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 0 which are divisible by 5 and none of the digits is repeated?

OPtion

1) 16
2) 18
3) 20
4) 24
5) 26
6) 28
7) 30
8) 32
9) 34
10)36

Solution

To be divisible by 5 unit digit of number should be either 5 or 0
We have total 6 digits out of which we cannot have 0 at the hundred’s place (else it will become a 2 digit number)
Unit place is filled with 5
a. Tens place can be filled with 0
Hundreds place can now be filled with (2, 3, 6, 7). We have 4 ways
So we have 1*1*4
b. Tens place can be filled with any of the digits (2, 3, 6, 7). We have 4 ways
Hundreds place can now be filled with remaining of 3 digits. We have 3 ways
So we have 1*4*3

Unit place is filled with 0
Tens place can be filled with any of the digits (2, 3, 5, 6, 7). We have 5 ways
Hundreds place can now be filled with remaining of 3 digits. We have 4 ways
So we have 1*5*4

So Total number of ways = 1*1*4 + 1*4*3 + 1*5*4 = 36

Option 10)

Correct Option: 10 Best Solution (14)

Nilesh   8 years AGO

Numbers ending with 5= 4x4=16
numbers ending with 0= 5x4=20
total = 36

Like? Yes (4) | No (3)  

Vaishali Jain   8 years AGO

0 at ones place : 4*5= 20 numbers
5 at ones place : 1) 0 at tens place : 4*1= 4
2) no 0 at any place : 3*4= 12
total number = 20+4+12 = 36

Like? Yes (2) | No (2)  

abhishek   8 years AGO

Case:1 Number Ending with 0 Then we have total 5 digit to fill in hundredth place and 4 digit to fill in tenth place (5*4)=20..

Case 2: ending with 5 then for hundred place we have 4 option because starting with 0 become two digit number.
Then for tenth place we have 4 choices.Totally become 4*4=16...

so total number of cases become 20+16=36 option 10)

Like? Yes (1) | No (1)  

MP    8 years AGO

3 digit Numbers ending with 0= 5*4=20
3 digit numbers ending with 5= 4*4=16
Total =20+16=36

Like? Yes (1) | No (2)  

Mukund Manohar   8 years AGO

Case I - When 0 is at the units place,
No. of ways to get 3-digit numbers = 4*5*1 = 20
Case II - When 5 is at the units place
No. of ways to get 3-digit numbers = 4*4*1 = 16
So,required 3-digit numbers = 20 + 16 = 36

Like? Yes (1) | No (2)  

G.S.Kantharao   8 years AGO

Those are 205,230,235,250,260,265,
270,275,305,320,325,350,
360,365,370,375,520,530,
560,570,605,620,625,630,
635,650,670,675,705,720,
725,730,735,750,760,765
Totally 36.
Hence the right option is 10.

Like? Yes (1) | No (2)  

Siva   8 years AGO

Given digits are 2,3,5,6,7 and 0.
From these digits when we form a three digit number which is divisible by 5 then there are only two ways to fill the unit place, either with 5 or 0.
1.If put 0 in unit place, then there are 5digits and two places should be filled with the 5 digits that to not repeat. So the number of ways are 5×4= 20.
2. If unit place is filled with 5, then we have 5digits and two places. But 0 must not be at 100s place so the 100s place will fill in 4ways and we have left with 4digits and the number of ways to fill 10s place with these 4 digits is 4. So totally in this method 4×4=16 ways.
So totally the number of ways required is 20+16=36

Like? Yes | No (2)  

Mini Agarwal   8 years AGO

5 options for hundredth's place(all digits excluding zero)
5 options for tenth's place(all digits except one that was selected for hundredth's)
total options=5*5=25
25 for 0
25 for 5
at tenth's place.
Total cases=25+25=50
Cases having repeated digits=255,355,655,755,525,535,565,575,505,200,300,600,700,500=14 cases
Total cases=50-14=36

Like? Yes | No (2)  

Jebas Gnana Prakash E   8 years AGO

All numbers except 0,5 form 8 count and 5 form 4 count
So the result is (4*8)+4=36

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Mathi Vanan P   8 years AGO

All no except 0,5 form 32 numbers and 5 form 4 numbers
So the result is 32+4=36

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lingadurai   8 years AGO

all no except 5,0 form 32 different number and 5 can form 4 numbers
totally
the result is(32+4)=36

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sandhip kumar padhy   8 years AGO

first use 0 in first place it become 20 then use 5 it become 16.add both.

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Testing Account   8 years AGO

**0
above is the 3 digit no.
at first position can be any no.except 0 otherwise it will be 2 digit no.
so for this 5*4=20combination.
and for
**5
4*4=16
hence total=20+16=36

Like? Yes | No (2)  

Vivek Yadav   8 years AGO

case 1: 0 is at last position:
then rest two places can be filled by rest 5 digits = 5*4 = 20

case 2: 5 is at last position
Then rest two places can be filled by rest 5 digits = 5*4 = 20
But this contains the result where 0 is at first place.
Total number when 0 will be at first place
0_5 this filled can be filled with 2,3,6,7 = 4 ways

So for case 2 total number 20 - 4 = 16

Hence the answer from case 1 and case 2 = 20 + 16 = 36

Like? Yes | No (2)