Find out the distinct numbers that can be formed by 2, 3,
7, 6(don?t now exactly) that should be divided by 4?
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Answer / bagesh kumar bagi
possible number is 8.
numbers are
3276,2376,7632,6732,6372,3672,7236,2736
these r possible number.
| Is This Answer Correct ? | 21 Yes | 0 No |
Answer / subbu
for a number to be divisible by 4, last two digits must be
divisble by 4.
so 32,36,72,76 are the only for possible combinations at
the last two positions.
The first position can be filled in 4 ways and similarly
second position also can be filled in 4 ways. so totally in
4*4=16 way for filling firt two positions.
So totally 16+4=20 distinct numbers can be formed
| Is This Answer Correct ? | 6 Yes | 5 No |
Answer / krishnapriya s
from the given digits ,the numbers ending with the
following combinations are divisible by 4, ie 76,72,36 and
32( so 4 chances)
first and second positions can be filled by the remaining
two didgits in two ways each..ie 2*2=4
hence the total distinct numbers that can be obtained is
4*4=16
| Is This Answer Correct ? | 4 Yes | 5 No |
Answer / lajapathy
last two numbers are any one of 32,36,72,76
first digit 4C1 second digit 4C1
so 4C1 X 4C1 X 4(four combinations of last two digits)
= 64
64 possible ways
| Is This Answer Correct ? | 7 Yes | 8 No |
Answer / priya
considering the different numbers with
23 (67/76) - 2 numbers
26 (37/73) - 2 numbers
27 (36/63) - 2 numbers
totally 6 numbers
In the same way 6 numbers can be formed for the rest 3
numbers (3,7,6), so totally 24 distinct numbers can be
formed.
| Is This Answer Correct ? | 3 Yes | 7 No |
1.BC CE EG GK ? a)KN b)KU c)KM d)None 2.AA AB BC CE? a)EG b)EH c)EI d)None 3.AB EF JK QR ? a)YZ b)ZA c)AB d)None 4.ACD EGL IKT MOB? a)QST b)QSZ c)QSY d)None 5.AC CG GO OE? a)EJ b)EI c)EL d)None 6.AE BH CM DU? a)EH b)EZ c)EB d)None 7.AD DP PL LV a)VS b)VK c)VI d)None 8.SE QU EN TI? a)CN b)BM c)AI or AZ d)None
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