emmak wrote:

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?

None

1

2

8

9

Constraint 1) when we put a sequence of multiples of (n+2) atleast one multiple should have its unit digit same as that of (n+1)

Constraint 2) N+1 can not greater than 9 since it is a single digit.

Constraint 2) N can not be 0

n(n+1)(n+2)

8---9----10------ Units digit zero always. so out

7---8----9-------- 9X2=18 so 2323232X9 = .......2988 or 2222222 x 9 = ......98 out

6---7----8-------- units digits 8,6,4,2,4,8,6,4,2,0,8,6.... No 7 so out

5---6----7-------- 7x8=56 so 83838 x 7 = .......5866 or 88888 x 7 = ....216 out

4---5----6-------- units digits 6,2,8,4,0,6,2,8,4,0...... No 5 so out

3---4----5-------- 5,0,5,0 out

2---3----4-------- 4,8,2,6,0,4.... out

1---2----3-------- only possible pair is 3 x 4 = 12 so out

Bunuel, Can you Pls help?

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