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19 Apr 2019, 08:59
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
53% (02:58) correct
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How many 4 digit numbers greater than 4000 can be formed using the digits from 0 to 8 such that the number is divisible by 4?
A. 508
B. 827
C. 828
D. 1034
E. 1035
A. 508
B. 827
C. 828
D. 1034
E. 1035
20 Apr 2019, 00:44
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mangamma
Multiple of 4 depends on last two digits.. a,b,x,y
a can be anything from 4 to 8, so 5 ways
b can be any of 9 digits (0 to 8)
1) If x is odd (1,3,5,7), y has to be non multiple of 4, (2,6)..
Total 5*9*4*2=360
2) If x is even , (,2,4,6,8), then y has to be multiple of 4, (0,4,8)
Total 5*9*5*3=45*15=675
Total = 360+675=1035. But this includes 4000, so 1035-1=1034
D
A quick answer with a bit of logic
1)The hundreds digit can have any of the nine digits..
_,(0-8),_,_
So the total should be a multiple of 9 if 4000 is also inclusive.
But here we are looking at >than 4000. So the total should be 9x-1.
2) The thousands digit can be any of 5 digits 4 to 8
(4-8),_,_,_
So total should be a multiple of 5 but as we are not including 4000.
Thus 5y-1
So our answer will be of the type 45z-1..
Clearly the units digit will be (5-1) or (10-1)
Only D fits in.
20 Apr 2019, 00:57
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Let a 4 digit number which is greater than 4000 and divisible by 4 is 'abcd'
'a' can take 5 values (4,5,6,7,8)
'b' can take 9 values (0,1,2,3,4,5,6,7,8)
If a number is divisible by 4, its last 2-digits must be divisible by 4
Hence 'cd' can take 23 values(00,04,08......88)
Total combinations possible=(5*9*23)-1=1034 (we deduct 1 because we have to exclude 4000)
'a' can take 5 values (4,5,6,7,8)
'b' can take 9 values (0,1,2,3,4,5,6,7,8)
If a number is divisible by 4, its last 2-digits must be divisible by 4
Hence 'cd' can take 23 values(00,04,08......88)
Total combinations possible=(5*9*23)-1=1034 (we deduct 1 because we have to exclude 4000)
