How many four digit numbers that are divisible by 4 can be formed usin

Question

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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ScottTargetTestPrep · Accepted Answer

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

nick1816 · Answer

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

Archit3110 · Answer

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

ism1994 · Answer

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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How many four digit numbers that are divisible by 4 can be formed usin

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