#50 Puzzle-by-numbers
Set by Matthew Scroggs and TD Dang
Solve these six puzzles, then find each region of the image below that contains one of the answers and colour it brown. Each answer may appear more than once.
1. Holly added up the first four odd numbers. What total did she get?
2. Holly added up the first m odd numbers to get a total of 441. What is m?
3. The number n! is calculated by multiplying together all the integers from 1 to n. How many zeros does the number 10! end with?
4. How many zeros does the number 50! end with?
5. How many three-digit, positive integers are there whose digits are all 1, 2, 3 or 4, with exactly two digits that are 1s?
6. Ivy works out that there are 72 n-digit positive integers whose digits are all 1, 2, 3 or 4, with exactly n-1 digits that are 1s. What is n?
What Christmassy image have you created?
#51 Ribbon theory
set by Peter Rowlett
I have a large supply of ribbon in lengths of 10 centimetres, 20 cm, 40 cm and 120 cm.
I could make 50 cm of ribbon by using one 40-cm piece and one 10-cm piece. How many other ways can I make 50 cm of ribbon? (Only the combination of lengths, and not the order of the pieces, matters.)
How many ways can I make 80 cm of ribbon?
If I am allowed to cut any one piece of ribbon in half, how many ways are there now to make 80 cm of ribbon?
#52 Tinsel theory
set by Phil Lloyd
Imagine two strands of tinsel tangled together. There are three ways of joining their four ends together in pairs. What is the probability that doing so will create a single, continuous loop?
What if there were three such tinsel strands, or six of them?
Can you find the formula for any number of tinsel strands?
#49 Squares and cubes
Solution
(set on 7 December)
The pair of numbers is 6 and 10. The difference in their squares, 10脳10 - 6脳6, is 64, which is 43. The difference in their cubes, 10脳10脳10 - 6脳6脳6, is 784, which is 282.
If this pair of numbers is multiplied by a number that is both a perfect square and a perfect cube, the resulting pair of numbers will have the same property. Any number raised to the power of 6 is both a perfect square and a perfect cube, as x6 = (x3)2 = (x2)3. The smallest such number is 26 or 64, so (6, 10) multiplied by 64 becomes (384, 640), which has the same property.
#50 Puzzle-by-numbers
Solution
The answers to the six puzzles are: 1) 16, 2) 21, 3) 2, 4) 12, 5) 9 and 6) 24. Colouring these regions brown will reveal an image of a gingerbread man.
#51 Ribbon theory
Solution
There are three further ways to make 50 cm of ribbon: 5脳10 cm, 3脳10 cm+1脳20 cm and 1脳10 cm+2脳20 cm. There are nine ways to make 80 cm of ribbon with no cuts: 8脳10 cm, 6脳10 cm+1脳20 cm, 4脳10 cm+2脳20 cm, 2脳10 cm+3脳20 cm, 4脳20 cm, 4脳10 cm +1脳40 cm, 2脳10 cm+1脳20 cm+1脳40 cm, 2脳20 cm+1脳40 cm and 2脳40 cm. Cutting a 120-cm piece in half to make a 60-cm piece available adds two new ways to make 80 cm. Cutting a 10-cm piece in half to make two 5-cm pieces adds another six ways. Cutting the 20-cm or 40-cm pieces in half adds nothing, since these sizes are already available. So, there are 17 ways to make 80 cm of ribbon with a single cut.
#52 Tinsel theory
Solution
If one tinsel strand has ends A and B, and the other C and D, then joining A to B (and C to D) gives two loops, but the other two combinations both give one loop. The chance of creating a single loop is therefore 2/3, or about 67 per cent.
With three strands of tinsel and six ends, choosing one will leave five free ends, of which four wouldn't immediately form a short loop. That gives a 4/5 chance of success. There are now four free ends (as above), so the overall chance of forming a single loop is 4/5 脳 2/3 = 8/15, or about 53 per cent. Following the same reasoning, with 12 free ends, the chance of forming a single loop is 10/11 脳 8/9 脳 6/7 脳 8/15 = 3840/10,395, or about 37 per cent.
The general formula for n tinsel strands is given by multiplying together (2n-2), (2n-4), (2n-6) and so on until one of the bracketed terms equals 2, then dividing by the corresponding odd terms starting from (2n-1):