11 Plus Exams Forum

Please can you help us with a simple technique to solve the following questions:

1. A Cuboid's faces have perimeters 10cm, 12cm and 14cm. Find its dimensions?

2. The Goldbach Conjecture states that every integer greater than 2 can be expressed(not uniquely) as the sum of two primes.
For example: 14=3+11 or 7+7
Find as many ways as you can in which 36 can be written as the sum of two primes?

3. A positive whole number less than 100 has remainder 2 when it is divided by 3, remainder 3 when divided by 4 and remainder 4 when divided by 5. What is the number?

Thanks
A

I'll start you off with hints.

1) The semi perimeters will be 5cm, 6cm and 7cm - the semi perimeter is the sum of two of the sides of that face.

2) What are the prime numbers? 2, 3, 5, 7, ... (should be known up to at least 20) then try adding

3) Think about the patterns -
remainder 2 when divided by 3 .... 5, 8, 11
remainder 3 when divided by 4 .... 7, 11, 15
remainder 4 when divided by 5 .... 9, 14, 19
can you see when they will all be the same number without listing them? What is the number that is one less than a multiple of 3, 4 and 5 ..

Thanks for your reply.

Sorry, I didn't get any of it. Please can you elaborate on all the answers?

Thanks
A

Guest 55 gave you some good starters, it's always best to try and work things out, it helps consolidate the knowledge - much better then just remebering by rote

Start with number 2

Once you have a list of the first few primes (numbers which have only 2 factors) then you can work out which ones added together make 36

For the first one, imagine a shoe box. One face will be a rectangle with two verticals and two horizontals. So half the perimeter will just be one vertical and one horizontal. Call these a and b. The next face round will also have a vertical and horizontal which is half the perimeter, but the vertical will be the same as before, so call these lengths a and c. The bottom of the box will then be a rectangle with lengths b and c.
a+b=5 and a+c=6, so the box is one cm longer than its width. You could use algebra, but the numbers are so small it would be easier to just see what fits.

The trick to doing the last one is to multiply the numbers together. You then have a number which can be divided into groups of 3, 4 or 5. Imagine we put it into groups of 5 then took one away from just one group. We would be left with a number which had several lots of 5 and one group of remainder 4. What would happen if we divided it instead into groups of 4 then took one away from one of the groups? The quick trick with the last one is to multiply all the numbers together then take one away.

I did give further hints by PM at the OP's request.

Hi There,

Was going thru these questions. Found them really interesting. Can anyone tell me where these questions have been picked up from???

I have not come across any such questions while preparing (Bond, CEM, GL , Letts)

Thanks
Jas

Just thought I'd point out that the Goldbach Conjecture is for every even number. (For an odd number, one of the primes must be 2, which leaves Hobson's choice on the other).