Tough but V.Good PS

[Vaibhav Sinha]

After multiplying a positive integer A, which has n digits, by (n+2), we get a number with (n+1) digits, all of whose digits are (n+1). How many instances of A exist?

A. None
B. 1
C. 2
D. 8
E. 9

OA: After some posts. I want to see how everyone approaches.

Source: Veritas Prep.

[Matt P]

Ok; so first I'd stick in a few numbers to get a better sense of the problem.

Say A has 3 digits. Multiply by 5, and get a number with 4 digits. All digits are equal to 4. So A becomes 4444.

Based on the "a number with (n+1) digits, all of whose digits are (n+1)", we have the following possible end values:

1, 22, 333, 4444, 55555, 666666, 7777777, 88888888, 99999999

We got to these numbers by multiply A by (n+2), so we can use a divisibility check to rule out options.

Does 1 divide 2? No

Does 22 divide 3? No

Does 333 divide 4? No

Does 4444 divide 5? No

Does 55555 divide 6? No

Does 666666 divide 7? Yes: 666666/7 = 95238 --> So A = 95238, n = 5, n+2 = 7, n+1 = 6.

Does 7777777 divide 8? No

Does 88888888 divide 9? No

Does 999999999 divide 10? No

(By A divides B, I mean is B a factor of A)

Ok, so we have one value. The answer is B.

[vaibhav]

Hi Matt,

Good to see you alive matey. :P Excellent solution. Yes, B is correct OA. 

There is a slight tweak that you can do to select the right (n+1) number. Any odd number of digits in (n+1) Number will have a even factor. This means dividing odd number by an even number ( as n+1 = odd ; n+2= even). Hence, definitely we can eliminate testing 333, 55555, 7777777, etc. This leaves us to test only the even set of numbers. Kinda reduces down the calculation.

Vaibhav.

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Regards,

Vaibhav

[Matt P]

Heh, yeah I tried to throw down a complete solution. In reality I wouldn't write down all of those numbers since as you said, odd numbers wouldn't have even factors, and some of the options can very obviously be cancelled out.