Solution
We have 0 + 2 + 3 + 4 + 6 = 15.
Since number is to be divisible by 6 meaning it need to be divisible by 2 and 3.
So, sum of digits need to be multiple of 3 and unit digit should be an even number.
Possible combination:
1. 2, 3, 4, 6
2. 0, 2, 3, 4
3. 0, 2, 4, 6
For 1st combination: Units place can be taken by 3 numbers, tens can be taken by 3, hundreds by 2 and thousand s place by 1 way.
Numbers possible in this
case = 1 x 2 x 3 x 3 = 18
For 2nd combination: Total number of numbers possible without any condition = 4! = 24.
Now out of 24, 6 will have 3 at units place and hence to be eliminated.
6 numbers will have 0 at thousands place, hence need to be eliminated.
Out of 12 eliminated numbers, 2 numbers which are 0243 and 0423 are deleted twice and hence need to be added.
Therefore, numbers possible in this cases = 24 – 6 – 6 + 2 = 14.
For 3rd combination:
Total number of numbers possible without any condition = 4! = 24.
Out of 24, 6 will have 0 at thousands place and hence need to be eliminated.
Therefore, numbers possible in this cases = 24 – 6 = 18
Total numbers possible = 18+ 14 + 18 = 50.
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How many four digit numbers that are divisible by 4 can be formed usin [#permalink]
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How many four digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number?
A. 520
B. 432
C. 370
D. 353
E. 345
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Re: How many four digit numbers that are divisible by 4 can be formed usin [#permalink]
Unit digit must be even
CASE 1- unit digit is 2 or 6
tens digit must be odd. We have 4 choices
(1,3,5 or 7)
For thousands digit, we have 6-1=5 (can't be 0) choices
For hundreds digit, we have remaining 5 choices
Total possible numbers= 5*5*4*2=200
Case 2- unit digit is 0
tens digit must be even. We have 3 choices (2,4 or 6)
For thousands digit, we have 6 choices
For hundreds digit, we have remaining 5 choices
Total possible numbers= 6*5*3*1= 90
Case 3- unit digit is 4
tens digit must be even. We have 3 choices (0,2 or
6)
1. when tens digit is 0
For thousands digit, we have 6 choices
For hundreds digit, we have remaining 5 choices
total possible numbers= 6*5*1*1=30
2. when tens digit is 2 or 6
For thousands digit, we have 5 choices
For hundreds digit, we have remaining 5 choices
total possible numbers= 5*5*2*1=50
Total possible numbers in all cases= 200+90+30+50=370
Bunuel wrote:
How many four digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number?
A. 520
B. 432
C. 370
D. 353
E. 345
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Re: How many four digit numbers that are divisible by 4 can be formed
usin [#permalink]
total possible combinations
unit digit as 0 = 6*5*3*1 ; 90
unit digit as 2/6 ; and (1,3,5,7) in tens 5*5*4*2 =
200
unit digit as 4 and (2,6) in tens ; 5*5*2*1 = 50
unit digit as 4 and 0 in tens ; 6*5*1*1 = 30
total = 90+200+50+80 ; 370
IMO C
Bunuel wrote:
How many four digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number?
A. 520
B. 432
C. 370
D. 353
E. 345
Are You Up For the Challenge: 700 Level Questions
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 16176
Location: United States (CA)
Re: How many four digit numbers that are divisible by 4 can be formed usin
[#permalink]
Bunuel wrote:
How many four digit numbers that are divisible by 4 can be formed using the digits 0 to 7 if no digit is to occur more than once in each number?
A. 520
B. 432
C. 370
D. 353
E. 345
Are You Up For the Challenge: 700 Level Questions
In order for a number to be divisible by 4, the last two digits of the number must be divisible by 4. Since we can choose only digits 0 to 7 and no digits can repeat, the last two digits of the number must be 04, 12, 16, 20, 24, 32, 36, 40, 52, 56, 60, 64, 72 or 76. Let’s separate these numbers into two groups - those with the digit 0 and those without the digit 0.
With the digit 0: 04, 20, 40, 60
Since these are the last two digits of the number, there are 6 choices for the first digit and 5 choices for the second digit. Therefore, there are 6 x 5 x 4 = 120 such numbers if the last two digits have the digit 0.
Without the digit 0: 12, 16, 24, 32, 36, 52, 56, 64, 72, 76
Since these are the last two digits of the number and the first digit of the number can’t be 0, there are 5 choices for the first digit and 5 choices for the second digit. Therefore, there are 5 x 5 x 10 = 250 such numbers if the last two digits do not have the digit 0.
Therefore, there are a total of 120 + 250 = 370 such numbers.
Answer: C
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Joined: 08 Nov 2020
Posts: 12
Re: How many four digit numbers that are divisible by 4 can be formed usin [#permalink]
Hi all,
I reasoned as following, please let me know if it makes sense:
total possible combinations of 4-digit numbers out of non-repeating values from 0->7: 7*7*6*5=1470 (i.e., not counting zero as a first digit)
Now, 1470 takes into account odd and even values. I'm interested in values divisible by 4, so I will split in halves 1470 twice, i.e.: once to get only even values, and again to get only values divisible by 4.
This leads to something close to 370, so my approach seems to work!
Thanks all
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Re: How many four digit numbers that are divisible by 4 can be formed usin
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Re: How many four digit numbers that are divisible by 4 can be formed usin [#permalink]
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