swerve wrote:N is the smallest number which is divisible by 49 and has all its digit same. How many prime numbers are factors of N?
A. 3
B. 9
C. 4
D. 7
E. 5
The OA is E.
Please, can any expert explain this PS question for me? I can't get the correct answer. I need your help. Thanks.
The number divisible by 49 having all the digit same would be one of the following.
1. 111...
2. 222...
3. 333...
4. 444...
5. 555...
6. 666...
7. 777...
8. 888...
9. 999...
We have no idea how many digits the smallest number N would have. We can take out the factor from above and then analyze them.
1. 111... = 111...
2. 222... = 2*(111...)
3. 333...= 3*(111...)
4. 444...= 4*(111...)
5. 555...= 5*(111...)
6. 666...= 6*(111...)
7. 777...=
7*(111...)
8. 888...= 8*(111...)
9. 999...= 9*(111...)
We see that the smallest number N can be achieved from 777... =
7*(111...) since
7 is divisible by one of the factors of 49, i.e. 7. So we have to find out the smallest number which is of the form 111... that is divisible by 7.
An efficient approach is Hit and Trial. We see that 111 is not divisible by 7; so let's try with 1,111, but is also not divisible. Same fate with 11,111. We see that the smallest number divisible by 7 is 111,111. Thus, N = 777,777.
N = 777,777 = 7*(111,111) is divisible by 3, 7, 11, 111, and 111,111.
The number of prime numbers of N are Five.
The correct answer:
E
Hope this helps!
-Jay
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