Alan

Given the data you supply, your calculations are correct.

Deleted.

Here's a rough and ready way to tackle this:

Notice that if b = 2a and c = 3a the left hand side of the first equation becomes 4a² + 6a² + 3a² = 13a²

In other words it is a multiple of 13 and hence satisfies the first equation.

For this situation, if a, b and c are all positive integers less than 13, then a can only be one of 1, 2, 3 or 4.

Trying these in turn we find that only a = 3 (hence b = 6 and c = 9) satisfy the second and third equations.

Hence a + b + c  =  5 mod(13)  

As follows:

Let side OA (opposite angle B) be b, and side OB (opposite angle A) be a.

From the cosine rule we have:  \(1^2=a^2+b^2-2ab\frac{\sqrt3}{2}\)  remembering that \(cos(30\deg)=\frac{\sqrt3}{2}\).

Rewrite as a quadratic in a:  \(a^2-\sqrt3ba+b^2-1=0\)

Solve for a in terms of b:  \(a=\frac{\sqrt3}{2}b+\frac{\sqrt{(2+b)(2-b)}}{2}\)  (This solution will give the largest value of a)

Differentiate a wrt b to get:  \(\frac{da}{db}=\frac{\sqrt3}{2}-\frac{b}{2\sqrt{(2+b)(2-b)}}\)

Set this to zero and solve for b to get:  \(b=\sqrt3\)

Substitute this back into the expression for a above to get:  \(a=2\)

(Corrected for earlier brain fade!!!).

I have no idea how to do it using permutations and combinations either - that's why I listed the pseudo code!

Here is a "brute force and ignorance" piece of pseudo code to count the number of five-digit even integers that have a digit sum of 13:

Set n = 0                                       n is the counter

for k = 10000 : 2 : 99998              loop from 10000 to 99998 in steps of 2

       a = floor(k/10^4)

       t = k – a*10^4

       b = floor(t/10^3)

       t = t – b*10^3

       c = floor(t/10^2)

       t = t – c*10^2

       d = floor(t/10)

       t = t – d*10

       sum = a + b + c + d + t

       if sum ==13 then n = n+1

end for loop

display n

Here, symmetry is used to tackle questions 1 and 2:

1. Both the constraint and the function to be maximised are symmetrical in x, y and z, so we should expect them all to have the same value at the maximum point. Hence let y=x and z=x so the constraint becomes: 3x² = 1 That is x = 1/√3 so x+y+z = 3/√3 = √3

2. Similarly, symmetry suggests a and b should take the same value.   Each term in the function to be minimised is of the form k/x where k is a constant.  The rate of change of a term like this with respect to x is -k/x².  We'd like the rate of change of each term in the function to contribute equally to the minimum value, so we want  -1/a² = -4/c²  and -1/a² = -16/d² so that c = 2a and d = 4a (we already have b = a ) Putting these values in the constraint equation we get  a + a + 2a + 4a = 1  or a = 1/8 Hence b = 1/8,  c = 2/8 and d = 4/8, so 1/a + 1/b + 4/c + 16/d = 64 the minimum value.

"In this problem, a and b are positive integers.
When a is written in base 9, its last digit is 5.
When b is written in base 6, its last two digits are 53.
When a-b is written in base 3, what are its last two digits? Assume a-b is positive."

In base 10 the last two digits of a make:         a1*9 + 5

In base 10 the last two digits of b make:        5*6 + 3 = 33

In base 10 the last two digits of a-b make:    d1*3 + d0

So:  a1*9 + 5 - 33 = d1*3 + d0  or a1*9 - 28 = d1*3 + d0  where 0<= a1 <= 8 and 0<= d1,d0 <= 2

Hence possible values for a1*9 - 28 are:  -28    -19    -10    -1    8    17    26    35    44 (just letting a1 run from 0 to 8)

The maximum value of d1*3 + d0 is 8, the minimum value is 0, because 0<= d1,d0 <= 2, and a-b can’t be negative, so the only option is d1*3 + d0 = 8

which means that d1 = 2 and d0 = 2. 

i.e. the last two digits of a - b in base 3 are 22

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As follows:

The first just says the current will be the same through two sections in series. The second says the total current through a section comprised of two parallel components will be the sum of the currents through each of the parallel components. The third says the total resistance of two components in series is the sum of the individual resistances. The fourth says to find the total resistance of two components in parallel, add the reciprocal of the component resistances and then take the reciprocal of the result.