Suppose

is the number of possible combinations for a suitcase with a lock consisting of

wheels. If you added one more wheel onto the lock, there would only be 9 allowed possible digits you can use for the new wheel. This means the number of possible combinations for

wheels, or

is given recursively by the formula

starting with

(because you can start the combination with any one of the ten available digits 0 through 9).
For example, if the combination for a 3-wheel lock is 282, then a 4-wheel lock can be any one of 2820, 2821, 2823, ..., 2829 (nine possibilities depending on the second-to-last digit).
By substitution, you have

This means a lock with 55 wheels will have

possible combinations (a number with 53 digits).
When the difference between one and two numbers is two, the two numbers are -10 and -12.
Given that,
The difference between one and two numbers is two. Twelve more than six times the second is four times the first.
We have to find the number.
We know that,
The system of equation we get
x=y-2 ----->equation(1)
4x=12+6y ----->equation(2)
Substitute for x in the equation(2)
4(y-2)=12+6y
4y-8=12+6y
4y-6y=12+8
-2y=20
y=-10
Substitute y=-10 in equation(1)
x=-10-2
x=-12
Therefore, The two numbers area -10 and -12 when the difference between one and two numbers is two.
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An exponential behavior can be observed.
Let x be the number of years she has to wait.
100((1+0.08/12)^12x) = 150
Now solve for x.
log base 1.08/12 of 1.5 = 61.02 approx
12x=61.02
x=5.09 years