Puzzles

In the UK, eight coins are needed: 1p, 1p, 2p, 5p, 10p, 20p, 20p, 50p.

In the US, ten coins are needed: 1¢, 1¢, 1¢, 1¢, 5¢, 10¢, 10¢, 25¢, 25¢, 25¢.

In a far away country, the unit of currency is the #, which is split into 100@ (# is like £ or $; @ is like p or ¢).

Let C be the number of coins less than #1. Let P be the number of coins needed to make any value between 1@ and 99@. Which coins should be the country mint to minimise the value of P+C?

The smaller diagonal square is made up of a 1×1 square and four 1×2 right-angled triangles. Therefore its area is 5.

The line FM, and therefore the line DN, has gradient 2. The line JM, and therefore the line ON, has gradient -½. ON passes through (1,2) and DN passes through (2,0). Therefore, DN has equation \(y=2x-4\) and ON has equation \(y=\frac{5}{2}-\frac{1}{2}x\). These lines intersect at \((\frac{13}{5},\frac{6}{5})\), these are the co-ordinates of N. By the same method, the co-ordinates of P are \((-\frac{8}{5},-\frac{1}{5})\).

By Pythagoras' Theorem, The diagonal of the larger square is \(\frac{7\sqrt{10}}{5}\) and so the area of the larger square is 9.8.

6210001000 has 6 zeros, 2 ones, 1 two, 0 threes, 0 fours, 0 fives, 1 six, 0 sevens, 0 eights and 0 nines.

Are there any more numbers like this? Prove that you have them all.

Let \(G\) be the number of girls, and £\(x\) be the amount each girl was originally to receive. This means that there are \(19-G\) boys who were to receive £\(x+30\). The total scholarship is £1000, so:

This simplifies to:

\(G\) is less than 19, so the only integer solution of this equation is \(x=40\) and \(G=11\).

After the funds are redistributed, each girl therefore receives £48. Let \(B\) be the amount each boy will now receive. The total is still £1000, so:

Which can be solved to find \(B=59\).

Each girl receives £48 and each boy receives £59.

Mrs. Coldcream still feels that this is unfair. How much more should be added to each girl's scholarship (resizing the boys' accordingly) so that the girls get at least as much as the boys?

Let the first digit be \(a\) and the final digit be \(b\). The cost of the 72 turkeys (in pence) is \(10000a+6790+b\). This must be divisible by 72.

\(10000a+6790+b\) is divisible by 72, so \(64a+22+b\) is divisible by 72. This means that \(b\) must be even (as 64,22,72 are all even). Let \(b=2c\).

Dividing by two, we find that \(32a+11+c\) is divisible by 36. \(c\) must be odd, so that \(32a+11+c\) is even. Let \(c=2d+1\) and so \(b=4d+2\).

Dividing by two again, we find that \(16a+6+d\) is divisible by 18. \(d\) must be even, so that \(16a+6+d\) is even. Let \(d=2e\) and so \(b=8e+2\). But \(b\) is a single digit number, so \(e=0\), \(b=2\) and \(16a+6\) is divisible by 18.

Dividing by two yet again, we find that \(8a+3\) is divisible by 9. \(a\) must be divisible by 3. Let \(a=3f\).

Dividing by three, we find that \(8f+1\) is divisible by 3. \(8f+1=6f+2f+1\) so \(2f+1\) is divisible by 3. This will be true when \(f=1,4,7,10,...\). \(a\) must be a single digit, so \(f=1\) and \(a=3\). And so the price of the 72 turkeys is £367.92, and one turkey will cost £5.11.

Which numbers could 72 be replaced with in the original problem so that the problem still has a unique solution?

The goods train reverses into the siding, leaving three trucks, then reverses out of the way.

The passenger train attaches the three trucks to its front then reverses out of the way.

The goods train backs into the siding.

The passenger train drives past the siding.

The goods train drives forwards out of the way.

The passenger train reverses, then drives forward into the siding, leaving the three trucks. It then drives forward out of the way.

The goods train reverses into the siding, picking up the trucks. The trains have now passed each other.

1. If the goods train can drive forwards into the siding, how could it let the passenger train through?

2. If the goods train had 8 trucks, how could it let the passenger train through?

Let \(S\) be the area of the large square, \(T\) be the area of one of the large triangles, \(U\) be one of the red overlaps and V be the uncovered blue square. We can write $$S=4T-4U+V$$ as the area of the square is the total of the four triangles, take away the overlaps as they have been double counted, add the blue square as it has been missed.

We know that 4U=V, so

Therefore one of the triangles covers one quarter of the square.

Five congruent triangles are drawn in a regular pentagon. The total area which the triangles overlap (red) is equal to the area they don't cover (blue). What proportion of the area of the large pentagon does each triangle take up?

\(n\) congruent triangles are drawn in a regular \(n\) sided polygon. The total area which the triangles overlap is equal to the area they don't cover. What proportion of the area of the large \(n\) sided polygon does each triangle take up?

Add up the number of doors leaving each room; call the sum \(S\). As the number in each room is even, \(S\) will be even. Each interior door has been counted twice (as they can be seen in two rooms) and each exterior door has been counted once. Let \(I\) be the number of interior doors and \(E\) be the number of exterior doors. We have: $$S=2I+E$$ $$E=S-2I$$ \(S\) and \(2I\) are even, so \(E\) must be even.

If the number of doors in each room is odd, is the number of exterior doors odd or even?