manpreetsingh86
Question: On multiplying a positive integer N having n digits by (n + 2), we get a number with (n + 1) digits, all of whose digits are (n + 1). How many such N exist?
(A) None
(B) 1
(C) 2
(D) 8
(E) 9
And the answer to your question is 9.
As for this question, here is the solution:
Let’s take an example to understand the question better.
"On multiplying a positive integer N having n digits by (n + 2), we get a number with (n + 1) digits, all of whose digits are (n + 1)"
Example: Let n = 3
_ _ _ × 5 = 4444
The numbers on the right hand side of the equation can only be of the form 1, 22, 333, 4444, 55555, 666666, 7777777, 88888888, 999999999 which gives us a limited set of possible options.
Looking at the example, we can find the 3 digit number, N, by dividing 4444 by 5. But since 4444 is not divisible by 5, there is no such positive integer.
Similarly, all we have to do is, using divisibility rules, divide each of these 9 numbers and see if they are divisible by their respective (n + 2)s. 1 is not divisible by 2, 22 is not divisible by 3, 333 is not divisible by 4 and so on… An odd number will never be divisible by an even number so we don’t need to check for odd numbers. So we ignore 1, 33, 55555, 7777777, 999999999
We find only 666666 is divisible by 7 giving 95338 as N. So there is only one such N.
Answer (B)
This question is at Q50/51, even though the starting point is clear, there is a small set of values and calculations involved are almost zero.