When positive integer A, which has n digits, is multiplied by (n+2) , the product is a number with (n+1) digits, all of whose digits are (n+1) . How many instances of A exist?

  A = 95238 (only)

Step-by-step explanation:

The question is equivalent to asking what n-digit number consisting only of the digit n will be divisible by n+1. Of the numbers 1, 22, 333, 4444, ... 999999999, the only one is 666666, which is divisible by 7.

The 5-digit integer A is 666666/7 = 95,238. There is only one instance of A.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

7

step-by-step explanation:

the answer would be 7