# Puzzles

## 18 December

The expansion of \((x+y+z)^3\) is

$$x^3 + y^3 + z^3 + 3x^2y + 3x^2z + 3xy^2 + 3y^2z + 3xz^2 + 3yz^2 + 6xyz.$$

This has 10 terms.

Today's number is the number of terms in the expansion of \((x+y+z)^{26}\).

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In each term, the powers of \(x\), \(y\) and \(z\) must add to 26: the terms will be of the form \(x^iy^jz^{26-i-j}\).

If \(i=0\), there are 27 choices for \(j\) (0 to 26).
If \(i=1\), there are 26 choices for \(j\) (0 to 25).
If \(i=2\), there are 25 choices for \(j\) (0 to 24).
...
If \(i=26\), there is 1 choice for \(j\) (0).

Therefore the total number of terms is 27+26+25+...+1 = **378**.

## 10 December

For all values of \(x\), the function \(f(x)=ax+b\) satisfies

$$8x-8-x^2\leqslant f(x)\leqslant x^2.$$

What is \(f(65)\)?

Edit: The left-hand quadratic originally said \(8-8x-x^2\). This was a typo and has now been corrected.

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If we plot the two quadratics, we see that they meet at \(x=2\) where they both have gradient 4.

The line \(f(x)\) must therefore pass through the point \((2,4)\) and have gradient 4, so its equation is \(f(x)=4x-4\).

This means that \(f(65)\) is **256**.

## 7 December

The sum of the coefficients in the expansion of \((x+1)^5\) is 32. Today's number is the sum of the coefficients in the expansion of \((2x+1)^5\).

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The sum of the coefficients can be found by substituting \(x=1\) into the expression. Hence the sum of the coefficients in the expansion of \((2x+1)^5\) is
\(3^5\), or **243**.

## 18 December

There are 6 terms in the expansion of \((x+y+z)^2\):

$$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz$$

Today's number is number of terms in the expansion of \((x+y+z)^{16}\).

## 10 December

The equation \(x^2+1512x+414720=0\) has two integer solutions.

Today's number is the number of (positive or negative) integers \(b\) such that \(x^2+bx+414720=0\) has two integer solutions.

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If the equation has two integer solutions, then it can be written as \((x+\alpha)(x+\beta)\), where \(\alpha\) and \(\beta\) are integers.
Expanding this and setting it equal to \(x^2+bx+414720=0\), we find that \(b=\alpha+\beta\) and \(414720=\alpha\beta\).

414720 has 55 different pairs of factors. Each of these pairs leads to two values of \(b\) (a positive and a negative value). Therefore there are **110** values of \(b\) that give the equation two integer solutions.

## Powerful quadratics

Find all real solutions to

$$(x^2-7x+11)^{(x^2-11x+30)}=1.$$

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If \(x^2-7x+11=1\) or \(x^2-11x+30=0\), then this is one. The solutions to these are \(x=2,5,\) and \(6\).

It could also be one if \(x^2-7x+11=-1\) and \(x^2-11x+30\) is even. This happens when \(x=3\) or \(4\).

Therefore all the solutions to this are \(x=2,3,4,5\) or \(6\).

## Two tangents

Find a line which is tangent to the curve \(y=x^4-4x^3\) at 2 points.

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At \(x=a\), \(y=a^4-4a^2\) and \(\frac{dy}{dx}=4a^3-12a^2\). Therefore the equation of the tangent at \(x=a\) will be \(y=(4a^3-12a^2)x+8a^3-3a^4\).

Taking this away from \(y=x^4-4x^3\) gives \(y=x^4-4x^3-(4a^3-12a^2)x-8a^3+3a^4\). We can now look at where this curve is tangent to \(y=0\) and look for a value of \(a\) that makes it tangent at two points.

If this curve is tangent to the \(x\)-axis at \(x=b\), then it will have a repeated root at \(x=b\). We know it is tangent at \(x=a\), so dividing \(x^4-4x^3-(4a^3-12a^2)x-8a^3+3a^4\) by \((x-a)\) twice gives \(x^2+(2a-4)x+3a^2-8a\). We want this to have a repeated root, hence the discriminant, \((2a-4)^2-4(3a^2-8a)\), must be 0.

Solving this gives \(a=1\pm\sqrt3\).

Therefore the equation of the line is \(y=-8x-4\).

## A bit of Spanish

Each of the letters P, O, C, M, U and H represent a different digit from 0 to 9.

Which digit does each letter represent?

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POCO is 4595 and MUCHO is 68925.

#### Extension

The question could be written as \(POCO\times 15=MUCHO\).

For which values of \(n\) are the letters uniquely defined by \(POCO\times n = MUCHO\)?