Help with yr6 maths homework please

[justanotherquestion]

Yr 6 DS brought home a maths paper last night, fairly easy and he whizzed through it, but we are all stumped by the right way to go about the 'reasoning type questions'.

Any help very gratefully received:

A positive whole number less than 100 has a 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?

We know the answer, but would like the technique.

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

Again, we have the answer but would like to know the best way to go about it?

Did you do it by trial and error, or use some process? I'd write down exactly what you thought to get to the answer.
I'll give them a go now, and see where I get to

I'm struggling to do the first by anything other than trial and error.
Second I drew a cube, labeled the three sides a, b, c.
Then perimeters must be 2a+2b, 2b+2c, and 2c+2a giving the 10, 12,14. Half each, and substitute to get the2,3,4 answer out

OK, slightly logical way for first q.
Nos, under 100 dividable by 5, with r4 are
99, 94, 89,84 etc

Only odd nos can work, as you will never get r3 when divided by 4 with an even no.
Rule out e.g 99, 39 as divided exactly by 3.
Test remaining nos (19,29,49,59,79,89) for remainders when divided by 3 and 4.

Is there a neatercway. I wouldn't like to find solutions under 1000 by that method.

We only have the answers, as they were attached to the back!

With the first we thought about the rules of divisibility and then thought about prime numbers, but think we may be missing something?

With the second we did it by trial and error, but I think you are right with the algebra. Thank you.

Anyone any thoughts about the first one, other than trail and error?

Let x be the number to find. To find the range of x, start by doing (1x3)+2=5. The smallest number it could be is 5.

The largest number is two higher the number below 100 that is divisible by 3. 96/3=32, so the highest number is 98.

So x is somewhere between 5-98.

So we know that we're looking at a number that is 2 further on than the numbers in the three times table.

3 (5), 6 (8), 9 (11), 12 (14), 15 (17), 18 (20), 21 (23), 24 (26), 27 (29), 30 (32), ....

It is also 3 bigger than the numbers in the 4 times table

4 (7), 8 (11), 12 (15), 16 (19), 20 (23), 24 (27).....

It is also 4 bigger than the numbers in the 5 times table

5 (9), 10 (14), 15 (19), 20 (24), 25 (29), ...

Compare the numbers in the 5x + 2 column with the numbers in the other columns and when you have a number that is in all 3 columns, that's it.

59/3= 19 r2; 59/4= 14 r3; 59/5= 11 r4

oh x post. Thank you!

The perimeter

As above: 2(a + b) = 10; 2(b + c) = 14; 2(a + c)= 12. a+b=5, b+c=7, a+c=6

To substitute, rearrange a+b=5: b=5-a.

substitute b in b+c=7: 5-a+c=7
Rearrange a+c=6: c=6-a
Substitute c in 5-a+c=7: 5-a+6-a=7
Tidy up: 11-2a=7 -> 7+2a=11 -> 2a=4 -> a=2

Then use that info in the other equations:
a+b=5 -> 2+b=5 -> b=3
a+c=6 -> 2+c=6 -> c=4

a=2, b=3, c=4

For the first one you could use a number square and colour in the ones that match the pattern until you find one that matches all three.

There must also be a way to do it with algebra and substitution.

The way that I would approach the first one is to find the lowest common multiple of the three multipliers in the question - so here, the numbers are 3, 4 and 5. The lowest common multiple of these is 60. However, we know that the answer is one below this, as in each case the remainder is one less than the next multiple. Therefore the answer should be 59.

I approached the perimeter one in the same way as lougle.

Thank you all.

Neat, Lemon that's the trick, you need to spot that you are one short of a full multiplier each time, so its lowest common denominator minus 1.
Presumably 119 also works then, but is above 100 so not a valid answer.