2DMommy:

Please help with this q :


Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (eg 123123). The number formed will always be divisible by 7,11 and 13.
A) explain why
B) what are the other factors of any such number formed ?

Thks !

Hi 2DMommy,

A) 7x11x13 = 1001
1001x100 = 100100 (answer)
B) Other factors are 2^2 x 5^2

Hope the above is correct.

Cheers!
Jtutor

Rhyll:

Hi jtutor,

I was looking at the math question ,
Do you mind explain how to know the power for 2 and 3.
is it by testing eg. 2^2 x 3^2 is a perfect square as well as perfect cube, then follow by guessing the power.

Hi Rhyll,

Yes this question involves some \"guess & check\" technique.

To have a perfect cube, the power for 2 and 3 must be either 3, 6, 9, etc (multiples of 3). To have a perfect square, the power must be either 2, 4, 6, etc (multiples of 2).

Examples of perfect squares: 2^2 x 3^2, 2^2 x 3^4
Examples of perfect cubes: 2^3 x 3^3, 2^3 x 3^6
(Pls note that the powers for 2 and 3 need not be the same.)

Cheers!
Jtutor

lost boy:

Please help me answer this question.


Find the smallest number,p,such that p/2 is a perfect square and p/3 is a perfect cube

Hi,

p should be 2^3 x 3^4 = 648.

p/2 will give 2^2 x 3^4 which is a perfect square with powers that are multiples of 2.
p/3 will give 2^3 x 3^3 which is a perfect cube with powers that are multiples of 3.

Cheers!
Jtutor

insanePaPa:

Sec1 Qn:


A class has between 30 to 40 students. Each boy in the class brings 15 chocolates for a class party. The chocolates are shared equally among 20 girls and a teacher with no leftovers.
i) How many students are there in the class?
ii) How many chocolates does their teacher receive?
:thankyou:

Hi insanePapa,

Since there are between 30-40 students in the class, there should be between 10-20 boys (after deducting 20 girls).

If each boy brings 15 chocolates for the party, and the total chocolates are shared among 20 girls + 1 teacher = 21 people, we must find the number of boys such that when it is multiplied by 15 chocolates, the total number of chocolates can be divided equally among 21 people.

15 = 3 x 5
21= 3 x 7
Hence, possible values for the number of boys will be 7, 14, 21, etc.
Since we know that the value should be between 10-20, the number of boys should be 14.

i) Number of students in the class = 20 girls + 14 boys = 34.

ii) Each of the 20 girls & 1 teacher will receive 15 x 14 / 21 = 10 chocolates. Hence, answer is 10.

Cheers!
Jtutor

Hi nounou,


To solve for 1+2+3+…+2010+2011+2012, you can group them together as follow:
(1+2012)+(2+2011)+(3+2010)+…
There will be a total of 2012/2=1006 pairs.
Hence 1006 x 2013 = 2,025,078.

Hope it helps.

Cheers,
Jtutor

Hi jovially,


I’m back and have just forwarded you the compilation for your reference.
Hope you will find it useful for your child.

Cheers!
Jtutor

jovially:

Jtutor:

[quote=\"Herbie\"]Hi jtutor, pls check yr pm

Hi Herbie,

As requested, I have just sent the Past Years' O Level Maths Trend Analysis to your email address.

Hope you find them useful.

Cheers!
Jtutor

Hi Jtutor,

May I have a copy too? TIA.

Cheers[/quote]Hi jovially,
I'm currently overseas and will forward you a copy once I'm back. Pls pm me your email address.
Cheers,
Jtutor

Hi Parents,


I have sent the attachments to you as requested.

All the best!

Cheers,
Jtutor

Herbie:

Hi jtutor, pls check yr pm

Hi Herbie,

As requested, I have just sent the Past Years' O Level Maths Trend Analysis to your email address.

Hope you find them useful.

Cheers!
Jtutor

Divha:

Hi Jtutor,


Thanks for your kind advice. Will do kinda assessment over her weak areas for the subject.

I'll certainly get help from you for her. Thanks a lot for your encouraging words.

Hi Divha,

You are most welcome

After you have clarified her doubt over the weak areas, you can give her more practice on those areas. Once she builds up her confidence, you can also introduce other school exam papers for her to stretch her thinking and understanding as the questions are typically more challenging. If you encounter any tough questions, pls feel free to post them here and I will try to provide the solutions as soon as I can.

Good luck!