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form 8 digit numbers from 1,2,3,4,5 with repetition is allowed divisible by 4..
Read Solution (Total 10)
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- no. is divisible by 4 if the no. formed by last two digits is divisible by 4.
so last 2 digits must be
12
32
52
24
44
remaining 1st six digits can be arranged in 5^6 ways.
therefore answer is 5^6+5^6+5^6+5^6+5^6 = 78125 - 11 years agoHelpfull: Yes(37) No(5)
- when repetition is allowed.
the number is divisible by 4 when its last two digit is divisible by 4
i.e. last two digits can be 12,24,32,44,52
NOW
if last two digit is 12 then 5^6 numbers are possible.
hence for 24,32,44,52
(five ways total)
answer is 5^7 78125 - 11 years agoHelpfull: Yes(15) No(4)
- no. is divisible by 4 if the no. formed by last two digits is divisible by 4.
so last 2 digits must be
12
32
52
24
44
now there are 8 places _ _ _ _ _ _ _ _
Last two digit can be filled in 5 ways
Since repetition is allowed 1st place can be filled in 5 ways, second place can also be filled in 5 ways....same for other 4 places.....so 5^6*5=78125 is the answer.
remaining 1st six digits can be arranged in - 11 years agoHelpfull: Yes(4) No(1)
- is it 5^7?
- 11 years agoHelpfull: Yes(3) No(4)
- 5^6*5=78125.
- 11 years agoHelpfull: Yes(3) No(1)
- no. is divisible by 4 if the no. formed by last two digits is divisible by 4.
so last 2 digits must be
12
32
52
24
44
remaining 1st six digits can be arranged in 5^6 ways.
therefore answer is 5^6+5^6+5^6+5^6+5^6 = 78125 - 11 years agoHelpfull: Yes(2) No(0)
- 62500 ????
- 11 years agoHelpfull: Yes(0) No(7)
- the answer shuld me 62500.
- 11 years agoHelpfull: Yes(0) No(6)
- there will be too many nos , i guess question should be no. of 8 digit nos
just place last two digits so that sum of them is divisible by 4 and calculate no. of combinations - 11 years agoHelpfull: Yes(0) No(2)
- no. is divisible by 4 if the no. formed by last two digits is divisible by 4.
so last 2 digits must be
12
32
52
24
44
remaining 1st six digits can be arranged in 5^6 ways.
therefore answer is 5^6+5^6+5^6+5^6+5^6 = 78125 - 11 years agoHelpfull: Yes(0) No(0)
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