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Question
How many five digit numbers divisible by 3 can be formed using the digits 0,1,2,3,4,7 and 8 if each digit has to be used at most once.
How many five digit numbers divisible by 3 can be formed using the digits 0,1,2,3,4,7 and 8 if each digit has to be used at most once.
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Solution
The correct option is B 744
We've to make 5 digit numbers using given 7 digits excluding 2 digits every time.Now,0+1+2+3+4+7+8=25.As 25 mod 3=1, duplets to be excluded should have their sum mod 3=1.It's not much difficult to find such duplets. Without much trouble,we get :(0,1),(0,4),(0,7),(1,3),(2,8),(3,4),(3,7).Thus, there are such 7 duplets & hence 7 such groups of 5 digits.
Now, we find that out of 7 groups, 4 groups have 0 in them which isn't allowed to take 5th place ( otherwise a 4-digit number will be formed.).Total numbers formed by these 4 groups =4(5!−4!)=384.
Other 3 groups have no zero so total numbers formed by them=3(5!)=360
Total possibel numbers 384+360=744
We've to make 5 digit numbers using given 7 digits excluding 2 digits every time.
Now,0+1+2+3+4+7+8=25.
As 25 mod 3=1, duplets to be excluded should have their sum mod 3=1.
It's not much difficult to find such duplets. Without much trouble,
we get :(0,1),(0,4),(0,7),(1,3),(2,8),(3,4),(3,7).
Thus, there are such 7 duplets & hence 7 such groups of 5 digits.
Now, we find that out of 7 groups, 4 groups have 0 in them which isn't allowed to take 5th place ( otherwise a 4-digit number will be formed.).
Total numbers formed by these 4 groups =4(5!−4!)=384.
Other 3 groups have no zero so total numbers formed by them=3(5!)=360
Total possibel numbers 384+360=744
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