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by govind_raj_76 » Sun May 02, 2010 11:32 am
How many 4 digit numbers that are divisible by 2 can be formed so that only the two rightmost digits are even and the digit 0 cannnot appear more than once ?

A) 300

B) 400

C) 500

D) 600

E) 700

Please explain the best approach to solve this problem ?

Thanks,
Govind
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by clock60 » Sun May 02, 2010 11:58 am
govind_raj_76 wrote:How many 4 digit numbers that are divisible by 2 can be formed so that only the two rightmost digits are even and the digit 0 cannnot appear more than once ?

A) 300

B) 400

C) 500

D) 600

E) 700

Please explain the best approach to solve this problem ?

Thanks,
i got D-600, reasoning
if i got the problem right the desired number must look like
odd odd even even and 0 used only once
for me all numbers will end 0,2,4,6,8

odd odd even 0=5*5*4*1=100 (4 -as we chose from 2,4,6,8 and 0 is already chosen)
odd odd even 2=5*5*5*1=125 ( here we have 0,2,4,6,8)
odd odd even 4=5^3=125
odd odd even 6=125
odd odd even 8=125

total=100+125*4=600
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by iamseer » Sun May 02, 2010 12:23 pm
IMO answer D

_ _ _ _

5*5*(4*1 +1*4+4*4) =600

Now, since only last 2 digits are even, the first 2 digits are odd.
First digit can be selected in 5 ways (1,3,5,7,9)
Second digit can be selected in 5 ways ((1,3,5,7,9)

Since 0 is not to be repeated, it can be at ten's place, unit's place or nowhere

0 in unit's place - Ten's place can be selected in 4 ways (2,4,6,8)
0 in ten's place - Unit's place can be selected in 4 ways (2,4,6,8)
0 nowhere - Unit's place and Ten's place can both be selected in 4 ways each

HTH
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by susantaiitk » Mon May 03, 2010 7:45 am
first 2 places are odd [ 1,3,5,7,9] total ways = 5*5 = 25
last 2 places are even [ 0,2,4,6,8] total ways = 5*5 = 25

but 0 can't be more than once, hence last 2 digit 00 is avoided then total ways of last 2 places = 25 - 1= 24

total ways = 25*24 = 600
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by sumanr84 » Mon May 03, 2010 8:49 am
susantaiitk wrote:first 2 places are odd [ 1,3,5,7,9] total ways = 5*5 = 25
last 2 places are even [ 0,2,4,6,8] total ways = 5*5 = 25

but 0 can't be more than once, hence last 2 digit 00 is avoided then total ways of last 2 places = 25 - 1= 24

total ways = 25*24 = 600
Fastest...Simple..clear and best solution..;-)
I am on a break !!
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by govind_raj_76 » Thu May 06, 2010 6:31 pm
clock60 wrote:
govind_raj_76 wrote:How many 4 digit numbers that are divisible by 2 can be formed so that only the two rightmost digits are even and the digit 0 cannnot appear more than once ?

A) 300

B) 400

C) 500

D) 600

E) 700

Please explain the best approach to solve this problem ?

Thanks,
i got D-600, reasoning
if i got the problem right the desired number must look like
odd odd even even and 0 used only once
for me all numbers will end 0,2,4,6,8

odd odd even 0=5*5*4*1=100 (4 -as we chose from 2,4,6,8 and 0 is already chosen)
odd odd even 2=5*5*5*1=125 ( here we have 0,2,4,6,8)
odd odd even 4=5^3=125
odd odd even 6=125
odd odd even 8=125

total=100+125*4=600
Nice Explaination !!
Govind
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by debmalya_dutta » Thu May 06, 2010 7:39 pm
Is the question missing the portion which applies the constraint that the 2 left most digits should be odd digits ? Strictly based on the question , the 1000 and the 100's place can have both even and odd digits.. why have you guys considered only 5 ways
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