36 cannot, cos need to be more than 36. 42 not divisible by 4. Only possible answer left is 48.

It's a factors and multiples question.

Thanks. How can I explain to my p3 kid?

P3? Then listing of the multiples of 2,3 & 4 for now. Factors & multiples is a P4 topic

With the range (36-50) given, she can skip right to the relevant numbers and find the common numbers. She has to understand that no remainder means that it is exactly divisible by the number. So that's an important concept there. Eventually, when she has mastered her multiplication tables, u can introduce her to divisibility tests, probably in P4 or P5.

Thank you so much. I've returned most of what I've learnt to teachers. Lol Now I'm picking everything up again at their levels.

Got to list down the factors for the 3 numbers and you will notice that the their highest common multiple below 50 is the answer and that is 48

2 x 3 x 4 = 24...... If you multiply by 3 or 4 again it will be more than 50.... So multiply by 2 = 48.

Actually, Vincent, your method doesn't account for multiple 36. LCM is 2,3,4 is actually 12. Should be 2x2x3 if u use Ladder method, so it's multiples of 12, not 24.

Thanks Wai Yen, but the question is such that we don't have to go to 12, 24 or 36 as the number has to be more than 36. Getting 48 and "check back" will see if the answer fulfills all conditions. By multiply 2, 3 and 4 will ensure they are also divisible.

Ah, I see ur reasoning now. 😉 *my bad* Thought I better clarify, 'cos when I did the ladder method at first, I actually made a mistake and ended up with 24, and my son corrected me. 😝 Thought you might have made the same mistake. *heh* Good to discuss. Thanks!

Thank you so much, Edlynn Rose and Yueh Mei.

Total increased by 53-35=18. Average increased by 45-42= 3. So 18/3=6 boys

Thank you so much.

Qn1 does not seem to have an answer. Effectively looking for a "perfect" number. Only know all even numbers.

Qn2 means to look for the first 10 even "perfect" numbers, which adds up to a super huge number.

Lol....you are right mate!

May I know what level of maths is this? Does not seem to be under school curriculum.

Factors, multiples and number patterns if we look at it simplistically. But it can go as high as researching it for a Phd thesis to proof the existence or non existence of an odd perfect number.

A heuristic test for the tractability of q1 is by looking at how hopeless the task of adding up factors for odd numbers is. The first odd non-prime is 9. Its factors add to 7 (1 + 3 + 3), and lags 9 by 2. The next number, 15, is worse; 1+3+5=9 lags 15 buy 6. The next, 21, has the sum of its factors (1+3+7=11) lag it by 10.

Got some mathematical formula to find perfect numbers. Look at it, can prove it cant be odd number then solve q1

2^n-1(2n-1) is the formula to calculate even perfect numbers. The question is how to find a formula to calculate the pattern for n. If you can come out with a formula for that pattern, you can give a speech in Harvard, Oxford, Cambridge and Yale to all the old Maths professors there and they will be excited to hear your speech.

There is no concrete proof that odd perfect numbers doesn't exist, so the question is still valid.

I tink can be proven somehow. But i'm not in e field of maths to explain in details. Somehow is not possible one. It must have 2 as a factor then can possibly add up to the number itself. By that, it has to be an even number.

Again, i think this is beyond the level of the people in this grp

For q1 the no. Cannot be prime

Factor cannot be even

Brendan, if you can find the answer for (1) you would be the first in the world.

I think no.1 is 5625

The factors of 5625 excluding 5625 add to less than 5000.

Questions to ask would be how to work out areas of rectangles (length multiply by breadth), what length and breadth should the unknown rectangle have that would also work for the other known rectangles. Therefore, we can apply concept of common factors of 75, 125 and 210...

Thank you! But I think I figure out that I should use the common length and to take ratio then it's just one step

Hi Aaron Hee, can show ur working?

Okie coming up...

Tell the pupils that there is a common factor, the length... It's not easy to see

Kate, I couldn't agree more but the question was pitch at a P5 level so I thought it's viable too. That's a good thing about this group, many solutions..😃

This could actually be solved without any algebra at all...just observe that as you slice up the left side, you can extend your slices to the right side, too. To that end, the problem becomes more accessible for the younger ones.