If abc represents the number 100a + 10b +c, then a, b and c must be integers.

Set a + 5b + 15c = 100a + 10b + c

Manipulate this to get c in terms of a and b, then see what the minimum values of a and b must be for c to be the smallest possible positive integer.

Here's a helping hand:

The following graphs should help:

Notice that the numerator of each term cancels with the denominator of the next term, so that overall you are left with \(\frac{a}{2}=9\)

Hence a = 18. The denominator of each term is just 1 less than the numerator, so b = a-1 = 17.

I get \(\sqrt2\) as follows:

Two ways to do it are as follows:

I'll leave you to solve the equations.

Set \(y=Ae^{u\times x}\) and plug this into the ODE. You shoud get two values for u, so your solution can be written as

\(y=Ae^{u_1 \times x}+Be^{u_2 \times x}\)

Use your boundary conditions to find A and B.

Like so:

Yes, the polynomial is:

p(x) = (-29/40320)x^{5} + (3/160)x^{4} + (-1571/8064)x^{3} + (163/160)x^{2} + (-3081/1120)x + (27/8)

This is the result I get, using Mathcad:

A graph often helps:

As follows:

See the following:

When x = 0, y = 7 so 7 = a*0^2 + b*0 + c

When x = 3, y = 1, so 1 = a*3^2 + b*3 + c

When x = 6, y = 7, so 7 = a*6^2 + b*6 + c

You now have three simultaneous equations for the three unknowns. I'll leave you to solve them.