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I'm going to be in the UKMT Team Challenge in a few days and revising some questions used in the previous year. The questions are really bugging my mind. I know it may seem like a lot and quite easy but thanks if you help out, I will be grateful.

$Question$ $1)$

How many times does the digit $3$ appear when the whole numbers between $1$ and $150$ are written down?

My thoughts:

I wrote down all the numbers from $1$ to $150$ with the digit $3$ in them and I only got $12$ digits. Also, is there any other way to count them that is faster without writing them down.

$Question$ $2)$

A car’s milometer shows that the number of miles it has travelled is $15951$. This number is palindromic which means it reads the same forwards as backwards. The next time a palindromic number appears on the milometer is exactly two hours later. Find the average speed of the car during these two hours.

My thoughts:

What I did was changed the number $15951$ to $16961$ and found out the difference which was $1010$. The answer is somehow $55mph$. How?

$Question$ $3$

A cube is divided into $64$ identical smaller cubes. Seven of these smaller cubes are removed. The resulting shape has a volume of $1539$cm3. What is the surface area of the original cube?

My thoughts:

I have no idea how to do this.

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Apart from other errors, $1010$ miles in two hours means $505\,\text{mph}$"! –  Hagen von Eitzen Apr 24 at 16:21
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Answered: ideone.com/kqs7Kp –  Shahar Apr 24 at 22:36
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4 Answers 4

up vote 9 down vote accepted

The next palindrome is 16061, not 16961.

7 of 64 cubes were removed, leaving 57. 1539 / 57 = 27, so each small cube is 3x3x3cm. The big cube was divided into 4x4x4 cubes, so it was 12x12x12 cm. The surface is six sides times 144 cm squared, or 864 cm squared.

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+1 ... and 16061 is what you get when you add 55*2 to 15951. –  Patrick M Apr 24 at 20:18
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(Q1)

this number between 0..50 and 100..150 is the same, so count the numbers with 3's in 0..50 twice and 51..99 once.

The first one is

  • where 3 occurs as 1st digit (10 total)
  • where 3 is a 2nd digit (3, 13, etc, 43) - 5 total
  • we counted 33 twice, but it has $3$ twice also, so there is no overcount.
  • subtotal $10+5 = 15$

The second one is just where the 3 occurs as a second digit with $\{5,6,7,8,9\}$ as a first one, so 5 total (53, 63, etc).

The final total then is $2 \cdot 15 + 5 = 35$.

(Q2) Hint the next one is 16061 not 16961

(Q3) Hint The surface area of the cube with side $a$ is $S=6a^2$ and volume is $V=a^3$. You have to figure out how removing these cubes impacts the surface area, compute $S$, the surface area of the original cube and use above info to infer $a$ and $V$.

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"How often does the digit 3 appear" - in 33 and 133, it appears twice so you shouldn't have removed 2. –  gnasher729 Apr 24 at 16:18
    
@gnasher729 fixed, thanks –  gt6989b Apr 24 at 16:21
    
@AndréNicolas misread the problem, was counting numbers with 3 not actual occurrences. Fixed now. –  gt6989b Apr 24 at 16:22
    
wouldn't $3$ occur as a $1st$ digit $11$ times instead of $10$? @gnasher729. There is the digit $3$ and then from $30$ up-to $ 39$ –  PerfectNutter Apr 24 at 16:30
    
For the purpose of this problem, 03 is a more convenient representation than plain 3 –  gt6989b Apr 24 at 16:32
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(Q2) The next palindromic number is in fact 16061.

(Q3) In reality we want to find the volume of the $64$ smaller cubes. It says that the volume of $57$ smaller cubes is $1539\mathrm{cm}^{3}$, so we can find the volume of one small cube then multiply it by $64$ to find the volume $V$ of the original cube. Now we know the volume we can find the surface area. Recall that the volume of a cube with sides $a$ is $a^{3}$. So $a = (V)^{\frac{1}{3}}$. Then recall the surface area of a cube is $S = 6a^{2}$.

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As usual people explained several solutions. Normally this is useful, but I think in this case you'll have to agree that all of the solutions don't require any specialised knowledge, and are common sense, in hindsight. The real question you, the OP, should ask yourself, is: given one of the solutions here, how would you have discovered it by yourself?

Example of fast solution for Q1:

  1. Imagine that the numbers from 1 to 150 are written down one under the other. Now where would we expect 3s to appear?

    • This is one of the key heuristics in mathematics: instead of doing something, imagine it and see if you can guess some of its properties, and then prove them using a convincing argument.
  2. Some numbers have 1 digit, others 2, the rest 3. So we can treat them separately.

    • This is another important problem-solving heuristic: treat different things separately. (Or make them the same somehow, like @gt6989b suggests.)
  3. The group of numbers with 2 digits each, written down one under the other is like a table with N rows and 2 columns. Instead of seeing it as a bunch of rows, we can look at it as two columns.

    • This is another heuristic: if you have a well-behaved object (like a grid of numbers), it can be usually cut up in several different ways.
  4. Notice that in the left column, each of the digits 1 to 9 appears the same number of times each (because there are the same number of 20-somethings and 90-somethings etc.). And in the right column, each of the digits 0 to 9 appear the same number of times each.

    • It's useful to establish (1) what things there are, and (2) how many there are of each, if you can.
  5. We know that there are 90 rows in that 2xN table, because there are 99 numbers from 1 to 99, and we don't include the numbers 1 to 9.

    • If it's not perfectly clear how many things there are, trace back to how we got them. Usually a thing is the result of some operation on two or more things together. If you know how many there were before, and what happens when you operate on them, then you know how many there are now.
  6. So there are 90 things in the left column and 1/9th are '3's. And there are 90 things in the right column and 1/10th of them are '3's. So, from both columns, there are 90/9 + 90/10 = 10 + 9 = 19. So there are nineteen 3s in the two-digit numbers group.

Well, you can see how it proceeds from here.

Actually, people have been trying to understand these problem-solving heuristics since the 1960s, so they could program computers to solve problems. And they succeeded for simple problems like these, but then essentially failed when, instead of mathematical problems, where everything is exact and means only one thing, they started to attack problems involving spoken language, people's relationships and communication when they're aware of each other's points of view, what happens to physical objects when various actions are applied to them, and other common sense problems that humans solve very easily, but where the objects and actions are not exact and well-defined. In 50 years, AI hasn't been able to replicate common sense in a machine. Even though many of the people who worked on this problem and failed went on to achieve major successes in other fields.

Anyway, this is why the only way to learn problem-solving is practice. Every time you solve a problem, your heuristic machinery gets slightly improved. When solving the problem about numbers 1 to 150, you don't just end up with the solution to that problem, but with some new heuristics (i.e. rules about what to do and where is a good idea to apply it) that go beyond this problem, and which will be useful in a good percentage of other problems you'll meet.

You might wonder: can we make a big list of all heuristics? I think it can be useful to try, but I expect that the list would be enormous. It would take several books to record it, and that would make it useless, if you read it and tried to memorise one at a time. Therefore, I think that solving new problems is the most efficient way to learn problem-solving techniques. However, once you've solved a problem that you initially struggled with, it is a very good idea to review what heuristics you've used, and which ones you should have used, and the exact point in your reasoning where you could have done better.

And most importantly, the things I wrote might seem simple, but the way very complicated problems are solved is using the same simple heuristics like some of those listed above, but many many times, and often a large number of them are used summarily (i.e. using a theorem that has been proved, and jumping directly from its premises to its conclusions, even if it took mathematicians collectively a few hundred years of applying simple heuristics to work out).

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