Tutor MathsGuru: Ask me for your burning Maths questions!

[iFruit]

Vanilla Cake:

Q1


Q3
Tom walks up a staircase.
Each time he can either take one step or two steps.
How many ways are there for Tom to walk up a ten-step staircase?

This is a fibonacci series. It is explained in the math hub olympiad challenge thread.

so the number ways for n steps taken will be in this form.

1 2 3 5 8 13 21 34 55 89 144...

so for 10 steps = 89 ways

[Maths Hub]

Vanilla Cake:

Q2
2004 students arrange themselves in a row.
In the first round of counting, they number themselves
1,2,3,1,2,3,1,2,3,........ from left to right.
In the second round of counting, they number themselves
1,2,3,4,5,1,2,3,4,5,1,2,3,4,5........ from right to left.
Find the number of students whose sum of numbers in the first and second rounds of counting is 5.



Source: Asia Pacific Mathematical Olympiad for Primary Schools 2004.

Sorry, I could not find the given answers for the above questions. Your effort and time to provide worked solutions for them are appreciated.

The answer is 402.

First, 2004 is divisible by 3, but not by 5 (So we will count from the right)

123123123123123
543215432154321

The last 15 digits yield 3 columns with sum count of 5.
Because 2004 = 133 x 15 + 9, we will have 133 groups of 15 students in which every group has 3 students with sum count of 5.

The remaining 9 students also gives us 3 students with sum count of 5:
123123123
432154321

Hence, the number of students with sum count of 5 = 134 x 3 = 402

[Maths Hub]

Vanilla Cake:

Q4
Two points A and B are 1100 m apart.
Alice and Ben leave point A at the same time and travel to and fro along a straight road between A and B at uniform speeds. Alice and Ben travel at 60 m/min and 160 m/min respectively. They both stop after 40 minutes.
(i) At which meeting are they nearest to point B?
(ii) Find the nearest distance in metre.

Source: Asia Pacific Mathematical Olympiad for Primary Schools 2004.

(i) 2nd meeting
(ii) 100m from point B

Since both Alice and Ben started at point A, the first time they will meet each other again will be after both of them have traveled 2200m altogether.
eg.
A-------------------->|<-------------(continue)
B--------------------------------------(make a u-turn)

Hence the total length traveled by both of them is 2 x length = 2 x 1100 = 2200m.

The total speed of the two is 60m/min + 160m/min = 220m/min

They will meet each other the first time in 2200/220 = 10 mins

The next time they meet each other (the distance have to be 4x1100m = 4400m, 6 x 1100m = 6600m and 8x1100m= 8800m)

We stop at 8800m because that's the total distance both of them can travel in 40 mins.

The meeting points are listed below:
Alice|Ben|Total
600m|1600m|2200m
1200m|3200m|4400m
1800m|4800m|6600m
2400m|6400m|8800m

Hence, from the table, we can see that at the second meeting, they are nearest from B (100m away).

[Maths Hub]

iFruit:

Vanilla Cake:

Q1


Q3
Tom walks up a staircase.
Each time he can either take one step or two steps.
How many ways are there for Tom to walk up a ten-step staircase?

This is a fibonacci series. It is explained in the math hub olympiad challenge thread.

so the number ways for n steps taken will be in this form.

1 2 3 5 8 13 21 34 55 89 144...

so for 10 steps = 89 ways

If Tom can only take 1 step or 2 steps at a time,
No of ways to climb up a 1-step staircase: 1 way
No of ways to climb up a 2-steps staircase: 1 + 1 = 2 ways
3-steps staircase: 1+2 = 3 ways
4-steps staircase: 2+3 = 5 ways
5-steps staircase: 3+5 = 8 ways
6-steps staircase: 5+8 = 13 ways
7-steps staircase: 8+13 = 21 ways
8-steps staircase: 13+21 = 34 ways
9-steps staircase: 21+34 = 55 ways
10-steps staircase: 34+55 = 89 ways

This is because the boy has only 2 choices at first: either to take 1 or 2 steps case. From 10 steps, if he chooses 1 step, then the no of ways immediately reduced to 9-steps, if he chooses 2 steps, then the no of ways reduced to 8-steps' case. Hope this helps.

This is in fact the fibonacci series, as mentioned by iFruit.

All Parents/Students, you can find more Maths Olympiad Questions at our thread:
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=14953&postdays=0&postorder=asc&start=0

[iFruit]

Maths Hub:

Vanilla Cake:

Q2
2004 students arrange themselves in a row.
In the first round of counting, they number themselves
1,2,3,1,2,3,1,2,3,........ from left to right.
In the second round of counting, they number themselves
1,2,3,4,5,1,2,3,4,5,1,2,3,4,5........ from right to left.
Find the number of students whose sum of numbers in the first and second rounds of counting is 5.



Source: Asia Pacific Mathematical Olympiad for Primary Schools 2004.

Sorry, I could not find the given answers for the above questions. Your effort and time to provide worked solutions for them are appreciated.

The answer is 402.

First, 2004 is divisible by 3, but not by 5 (So we will count from the right)

123123123123123
543215432154321

The last 15 digits yield 3 columns with sum count of 5.
Because 2004 = 133 x 15 + 9, we will have 133 groups of 15 students in which every group has 3 students with sum count of 5.

The remaining 9 students also gives us 3 students with sum count of 5:
123123123
432154321

Hence, the number of students with sum count of 5 = 134 x 3 = 402

ah..right

[iFruit]

Maths Hub:

Vanilla Cake:

Q4
Two points A and B are 1100 m apart.
Alice and Ben leave point A at the same time and travel to and fro along a straight road between A and B at uniform speeds. Alice and Ben travel at 60 m/min and 160 m/min respectively. They both stop after 40 minutes.
(i) At which meeting are they nearest to point B?
(ii) Find the nearest distance in metre.

Source: Asia Pacific Mathematical Olympiad for Primary Schools 2004.

(i) 2nd meeting
(ii) 100m from point B

Since both Alice and Ben started at point A, the first time they will meet each other again will be after both of them have traveled 2200m altogether.
eg.
A-------------------->|<-------------(continue)
B--------------------------------------(make a u-turn)

Hence the total length traveled by both of them is 2 x length = 2 x 1100 = 2200m.

The total speed of the two is 60m/min + 160m/min = 220m/min

They will meet each other the first time in 2200/220 = 10 mins

The next time they meet each other (the distance have to be 4x1100m = 4400m, 6 x 1100m = 6600m and 8x1100m= 8800m)

We stop at 8800m because that's the total distance both of them can travel in 40 mins.

The meeting points are listed below:
Alice|Ben|Total
600m|1600m|2200m
1200m|3200m|4400m
1800m|4800m|6600m
2400m|6400m|8800m

Hence, from the table, we can see that at the second meeting, they are nearest from B (100m away).

The answer is correct but I think there should be 5 meeting points.

After the second meeting point, Alice is travelling towards A and Ben is traveling towards B, They meet again after 2 mins ( alice travels 120m, Ben travels 320m).

Then meeting times are 10, 10, 2, 8, 10 and meeting points 500, 100, 220, 700, 900m away from B.

[Maths Hub]

iFruit:

Maths Hub:

[quote=\"Vanilla Cake\"]

Q4
Two points A and B are 1100 m apart.
Alice and Ben leave point A at the same time and travel to and fro along a straight road between A and B at uniform speeds. Alice and Ben travel at 60 m/min and 160 m/min respectively. They both stop after 40 minutes.
(i) At which meeting are they nearest to point B?
(ii) Find the nearest distance in metre.

Source: Asia Pacific Mathematical Olympiad for Primary Schools 2004.

(i) 2nd meeting
(ii) 100m from point B

Since both Alice and Ben started at point A, the first time they will meet each other again will be after both of them have traveled 2200m altogether.
eg.
A-------------------->|<-------------(continue)
B--------------------------------------(make a u-turn)

Hence the total length traveled by both of them is 2 x length = 2 x 1100 = 2200m.

The total speed of the two is 60m/min + 160m/min = 220m/min

They will meet each other the first time in 2200/220 = 10 mins

The next time they meet each other (the distance have to be 4x1100m = 4400m, 6 x 1100m = 6600m and 8x1100m= 8800m)

We stop at 8800m because that's the total distance both of them can travel in 40 mins.

The meeting points are listed below:
Alice|Ben|Total
600m|1600m|2200m
1200m|3200m|4400m
1800m|4800m|6600m
2400m|6400m|8800m

Hence, from the table, we can see that at the second meeting, they are nearest from B (100m away).

The answer is correct but I think there should be 5 meeting points.

After the second meeting point, Alice is travelling towards A and Ben is traveling towards B, They meet again after 2 mins ( alice travels 120m, Ben travels 320m).

Then meeting times are 10, 10, 2, 8, 10 and meeting points 500, 100, 220, 700, 900m away from B.[/quote]Yes. That's also a meeting point, where both person are travelling in the same direction.

[Vanilla Cake]

Thks to Maths Hub and iFruit for your detailed and clear solutions. Sorry, some more questions from APMOPS 2005 that need your help.


Q2
How many whole numbers from 1 to 1000 can be expressed as the difference of the squares of two whole numbers?
[Note: 0 is a whole number.]

Q3
The following number is made up of all the digits of the whole numbers 1 to 2005.
12345678910111213141516..........20042005
Find the number of zeros in this number.

Q6
Allen , Benedict and Carl started at the same instant from the same point using the same route trying to overtake a fourth cyclist Donald traveling at a constant speed ahead of them. Allen and Benedict each took 10 hours and 2 hours respectively to overtake Donald.
Given that Allen , Benedict and Carl each cycled at the constant speed of 4 km/h, 5 km/h and 10 km/h respectively throughout the journey, find the time, in hours, that Carl took to overtake Donald.

Source: http://www.hci.sg/aphelion/apmops/2007/pdf/English/2005%20English%20IR.pdf.

Could you pls advise the correct answers for Q1, Q4 and Q5 too?
Thks and much appreciated for your help.

[iFruit]

Vanilla Cake:

Thks to Maths Hub and iFruit for your detailed and clear solutions. Sorry, some more questions from APMOPS 2005 that need your help.


Q2
How many whole numbers from 1 to 1000 can be expressed as the difference of the squares of two whole numbers?
[Note: 0 is a whole number.]

we know a² - b² can be written as (a+b)(a-b) i.e. product of the sum of the two numbers and difference of the two numbers.

1) Assume a, b are both even or both odd. Then a+b is even and a-b is even. so (a+b)(a-b) is a multiple of 4.

That means every multiple of 4 can be written as a difference of two squares.

2)Assume one of a,b is odd and the other even. Then a+b is odd and a-b is odd
so (a+b)(a-b) is a odd number. For example it can be written as

(a+b)(a-b) = 2n+1 where (a+b)=2n+1, (a-b) = 1 giving a = n+1, b =n

That is every odd number can be written as a diff of two squares.


So all even numbers that can be divided by 4 and all odd numbers can be expressed as a sum of squares. Only even numbers that can be divided by 2 but not by 4 can't be expressed in a such a manner.

so there are 500 even numbers between 1..1000 of which only 250 can be expressed divided by 2 but not by 4.

So 750 numbers can be expressed as difference of squares.

[iFruit]

Vanilla Cake:

Thks to Maths Hub and iFruit for your detailed and clear solutions. Sorry, some more questions from APMOPS 2005 that need your help.


Q3
The following number is made up of all the digits of the whole numbers 1 to 2005.
12345678910111213141516..........20042005
Find the number of zeros in this number.

Corrected following maths hub solution.. Thank Maths hub!

1) we know between 2000 .... 2005 there are 13 zeros

2) Now consider numbers 1000 ... 1999

The 1000's digit is always one so 0 can appear only in 100's, 10's or 1's places.

If there is only one 0, in those numbers, the other two places can vary between 1 to 9 giving 9x9 = 81 possibilities. Since there are three places we could have 3x81 single zeros between 1000...1999

Similarly for two zeros, we will have 3x9 times number 0s appearing = 27x2 0's

3 0's (number 1000 ) = 1 time = 3 0's

3) Consider numbers 100 to 999

we get 2x 81 numbers with one zero, 9 numbers with two zeros (18 zeros)

4) Consider numbers 10..99

we can have 0 in 1's digit place 9 times

So that total number of zeros = 13 + 3x81 + 27x2 + 3 + 9x2 + 2x81 +9 = 502