All About Math Olympiad Training & Questions

[turquoise]

justbehappy:

I happen to know them, they did not have any external/formal training when they were in primary school. Formal training for them starts when they go to sec sch. Schools like RI, RGS, NUSH and HCI have their own regular MO training in the school - could be conducted by seniors, teachers or external trainers. These few schools usually take majority of the top 30 positions in the MO competitions in Singapore. Then the good ones will be selected to join the jnr/snr/national team training conducted at NUS Maths Dept. These are all free training (by the schools and by NUS), no need to pay a cent

They are indeed gifted in math to be able to perform so well with no training.

[PowerPuff Mum]

turquoise:

They are indeed gifted in math to be able to perform so well with no training.

Some children are born with such ability without additional help or special training, they are born with that innate talent, that's why they are called gifted. As parents, we should provide them the opportunity to shine and provide exposure for them to do well in what they are good at.

It becomes a problem only when parents has unrealistic expectations and compare with other high ability children. We are all different. Children should be loved and appreciated in whatever form they come in.

[mummyloves]

Yeah, keen to know which level is this qn for? :? Not so easy eh...

[Dharma]

mummyloves:

Yeah, keen to know which level is this qn for? :? Not so easy eh...

Hi mummyloves,

Qn is pitched at P6 level. Taken from one of the early SMOPS papers, I think.

Need to have understanding of distance, time and speed relationship. Then use ratios to solve.

[Courage]

Need :?: for the following questions:


(1) If we write whole numbers starting from 1 to form a 1000-digit number N as follows: N=12345678910111213.... what is the unit digit of N?

(2) How many whole numbers from 1 to 2009 contain the digit \"2\"?

(3) Study the following pattern:

1=1: 1(1+1)/2=1
1+2=3; 2(2+1)/2=3
1+2+3=6; 3(3+1)/2=6
1+2+3+4=10; 4(4+1)/2=10
1+2+3+4+5=15; 5(5+1)/2=15

Find the value of 22+24+26+....+200.

TIA.

[Pen88n]

Let me try to get the answers:


1) Between 1 - 99, total # of digits 189 digits
Betweem 100 - 199, total # of digits 300
Between 200 - 299, total # of digits 300
Between 300 - 399, total # of digits 300

So you can see the 1000th digit falls within number "370" and it ends at digit 3.

2) From 1- 100, there are 19 numbers with digit 2 (10 from the "20 - 29" range and 1 each from the "1 - 90 exclude 20+" range)

For 200 - 299, there are 100 numbers with digit 2

From 1 - 1000, there are 100 + 9(19) = 271 numbers with digit 2.

From 1001 - 1999, there are 271 numbers with digit 2

From 2000 - 2009, there are 10 numbers with digit 2

From 1 - 2009, total 271+271+10 = 552 numbers with digit 2

3) 22+24+26+…+200 = 200(200+2)/2 = 20200

Please advise if answers are correct. Thanks

[Courage]

Pen88n, Thanks for your reply.


My working for Q3 is

2(11+12+13+…+100)
2(1+2+3+4+5+…+100)

Using the formula,
1+2+3+4+…+100 is 100(100+1)/2 =5050

thus, 2(5050-55)=9990 since 1+2+3…+10=55

[kiasiparent]

Pen88n:

Let me try to get the answers:


1) Between 1 - 99, total # of digits 189 digits
Betweem 100 - 199, total # of digits 300
Between 200 - 299, total # of digits 300
Between 300 - 399, total # of digits 300

So you can see the 1000th digit falls within number \"370\" and it ends at digit 3.

2) From 1- 100, there are 19 numbers with digit 2 (10 from the \"20 - 29\" range and 1 each from the \"1 - 90 exclude 20+\" range)

For 200 - 299, there are 100 numbers with digit 2

From 1 - 1000, there are 100 + 9(19) = 271 numbers with digit 2.

From 1001 - 1999, there are 271 numbers with digit 2

From 2000 - 2009, there are 10 numbers with digit 2

From 1 - 2009, total 271+271+10 = 552 numbers with digit 2

3) 22+24+26+....+200 = 200(200+2)/2 = 20200

Please advise if answers are correct. Thanks

First 2 are correct
3) 45 pairs x 222 = 9990

[Pen88n]

Thanks, Courage & Kiasiparent.


Finally see \"light\" for Q3.

It's great workout for my brain

[Tang]

Pen88n:

Let me try to get the answers:


1) Between 1 - 99, total # of digits 189 digits
Betweem 100 - 199, total # of digits 300
Between 200 - 299, total # of digits 300
Between 300 - 399, total # of digits 300

So you can see the 1000th digit falls within number \"370\" and it ends at digit 3.

2) From 1- 100, there are 19 numbers with digit 2 (10 from the \"20 - 29\" range and 1 each from the \"1 - 90 exclude 20+\" range)

For 200 - 299, there are 100 numbers with digit 2

From 1 - 1000, there are 100 + 9(19) = 271 numbers with digit 2.

From 1001 - 1999, there are 271 numbers with digit 2

From 2000 - 2009, there are 10 numbers with digit 2

From 1 - 2009, total 271+271+10 = 552 numbers with digit 2

3) 22+24+26+....+200 = 200(200+2)/2 = 20200

Please advise if answers are correct. Thanks

1)
1 to 9 --> 9 digits
10 to 99 --> 2 x 90 = 180 digits
100 to 299 --> 3 x 200 = 600 digits

1000 - 9 - 180 - 600 = 211 digits

211 / 3 = 70 r 1 --> So number from 300 to 369 and 3(70) is incomplete
Hence the digit is 3.

3)
(2 + 4 + .... 200) - (2 + 4 + ....20)
= (2+200)/2 x 100 - (2+20)/2 x 10
= 10100 - 110
= 9990