Australian maths competition (1 Viewer)

[Mongoose528]

Alright, so the AMC is coming up soon and I would like some help on the number theory and geometry

It's questions like these that I just don't understand

The Fibonacci numbers are F1=1, F2=1, F3=2, F4=3, F5=5, F6=8, F7=13, …… where the first two are both equal to 1, and from then on, each one is the sum of the two preceding it. Of the first 2004 Fibonacci numbers, how many have 2 as their last digit?

and

The number of positive integers less than 300 000 that have the digits 4, 5 and 6 together and in that order is?

There are also difficult geometry questions that I don't understand

Any tips/advice would be highy appreciated

And does anyone know the cut-offs for prizes/HD's for yr 9 intermediate in the past few years?

 

[InteGrand]

Alright, so the AMC is coming up soon and I would like some help on the number theory and geometry

It's questions like these that I just don't understand

The Fibonacci numbers are F1=1, F2=1, F3=2, F4=3, F5=5, F6=8, F7=13, …… where the first two are both equal to 1, and from then on, each one is the sum of the two preceding it. Of the first 2004 Fibonacci numbers, how many have 2 as their last digit?

and

The number of positive integers less than 300 000 that have the digits 4, 5 and 6 together and in that order is?

There are also difficult geometry questions that I don't understand

For the second one, we can break it up into cases. The numbers we want must have at least three digits, and can have at most six digits (as 300 000 has six digits).

Three digits
Only possibility is clearly 456. So 1 possibility.

Four digits
The numbers must be of the form:

• 456 _ ; 10 possibilities (anything for last digit)
• _ 456 ; 9 possibilities (can't start with 0).

These two categories are clearly disjoint (different second digits).

Five digits
The numbers must be of the form:

• _ _ 456 ; 9×10 = 90 possibilities (can't start with 0 for first digit, and can have anything in second digit's place)
• _ 456 _ ; 90 possibilities
• 456 _ _ ; 10×10 = 100 possibilities.

We can see the three categories are pairwise and mutually disjoint as the third digit is different in each category.

Six digits (< 300 000)
The numbers must be of the form (noting we can't start with 456 since we must stay under 300 000):

• _ 456 _ _ ; 2×10×10 = 200 possibilities (first digit can only be from {1,2}).
• _ _ 456 _ ; similarly, 200 possibilities
• _ _ _ 456 ; Similarly, 200 possibilities.

We can see the three categories are pairwise and mutually disjoint as the fourth digit is different in each category.

So adding up, the total number of possibilities is 1 + 10 + 9 + 90 + 100 + 90 + 200 + 200 + 200 = 900.

 

[Mongoose528]

For the second one, we can break it up into cases. The numbers we want must have at least three digits, and can have at most six digits (as 300 000 has six digits).

Three digits
Only possibility is clearly 456. So 1 possibility.

Four digits
The numbers must be of the form:

• 456 _ ; 10 possibilities (anything for last digit)
• _ 456 ; 9 possibilities (can't start with 0).

These two categories are clearly disjoint (different second digits).

Five digits
The numbers must be of the form:

• _ _ 456 ; 9×10 = 90 possibilities (can't start with 0 for first digit, and can have anything in second digit's place)
• _ 456 _ ; 90 possibilities
• 456 _ _ ; 10×10 = 100 possibilities.

We can see the three categories are pairwise and mutually disjoint as the third digit is different in each category.

Six digits (< 300 000)
The numbers must be of the form (noting we can't start with 456 since we must stay under 300 000):

• _ 456 _ _ ; 2×10×10 = 200 possibilities (first digit can only be from {1,2}).
• _ _ 456 _ ; similarly, 200 possibilities
• _ _ _ 456 ; Similarly, 200 possibilities.

We can see the three categories are pairwise and mutually disjoint as the fourth digit is different in each category.

So adding up, the total number of possibilities is 1 + 10 + 9 + 90 + 100 + 90 + 200 + 200 + 200 = 900.

thanks, is this the method you usually use in solving these type of questions?

 

[InteGrand]

For the first, let's consider the sequence of "Fibonacci last digits". This is obtained by just taking the last two numbers in the sequence, adding them, and only recording the last digit. So it is:

1, 1, 2, 3, 5, 8, 3 (only record last digit of 13), 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, 1, 6, 7, 3, 0, 3, 3, 6, 9, 5, 4, 9, 3, 2, 5, 7, 2, 9, 1, 0, | 1, 1, 2, and the pattern repeats from the start.

So from the first 1 to the last digit (0) before the |, we will have our cycle, which repeats thereafter. We can count that there are 60 numbers in this cycle, with four 2's appearing.

So in the first 1980 = 33×60 numbers in the list (i.e. 33 cycles), we have 33×4 = 132 2's appearing. After this, we have 24 more numbers in the list of first 2004 numbers. We can see that in our cycle, in the first 24 numbers, the number 2 appears exactly once. So in the last 24 elements of the list of first 2004 numbers, the number 2 appears exactly once.

So in total, the number 2 appears 133 times in the first 2004 "Fibonacci last digits". So the answer is 133.

 

[Mongoose528]

thanks^

 

[Mongoose528]

Does anyone know what the prize and HD cut offs for the year 9 intermedaite division in the past 5 years were?

 

[RealiseNothing]

Here's a nice approach to question 2 which highlights a neat way of thinking about questions like this:

Consider an arbitrary 6 digit number:

A B C D E F

We must have 456 in here somewhere. So we already know 3 of the digits, meaning we really only have this:

_A_B_C_

Now we need to insert the 456 into our number. Notice 456 can not be the first 3 digits otherwise we exceed 300,000. So 456 can only be inserted into the 2nd, 3rd, and 4th spaces. So there are 3 ways of inserting 456.

Also for us to not exceed 300,000 we must have that the number ABC is between 0 and 299, i.e. 300 possibilities for ABC.

Hence the answer is 300*3=900 (# of possibilities of ABC times the number of ways of inserting 456 into our number).

This method deals with all cases simultaneously (i.e. if ABC=047 then we are have a 5 digit number which is trivially less than 300,000).

 

[Mongoose528]

thanks