Maths Question Help

[nikku]

Hi All

Please help with the fastest method of working out the following

1. Alison has large number of pencils. When the pencils are divided by two of her friends there is one pencil left over. If they are divided by three friends there are two left over. Divided between four friends there are three left over. Between give friends there are four left over and between six friends, there are five left over. What is the smallest number of pencils that Alison could have?

2. How many three digit numbers contain at least two sevens.

3. The number 3 can be split in three different ways by adding positive whole numbers together as follows
1+2, 2+1, and 1+1+1

Using the same method, in how many different ways can the number 5 be split?

4. There are 4 beads on a necklace, 2 beads are red, 1 bead is green and 1 bead is blue. How many different colour arrangements can be made from the beads?

Thank you

[Guest55]

Two of these (3 and 4) are about 'finding all the ways' which is a technique.

3. The number 3 can be split in three different ways by adding positive whole numbers together as follows
1+2, 2+1, and 1+1+1

Using the same method, in how many different ways can the number 5 be split?

So for 5:

1 + 1 + 1 + 1 + 1

1 + 1 + 1 + 2
1 + 1 + 2 + 1
1 + 2 + 1 + 1
2 + 1 + 1 + 1 by 'moving the 2' along we get all the ways

repeat for 2, 2 , 1

4. There are 4 beads on a necklace, 2 beads are red, 1 bead is green and 1 bead is blue. How many different colour arrangements can be made from the beads?

call the beads R, R, G and B

GBRR
GRBR
GRRB etc

1) ignore the pencils and look at the numbers - we want a number that is one less than a multiple of 2, 3, 4, 5 and 6

2) Again work logically
100 to 200 only one 177
200 to 300
.
.
700 to 800 more 717, 727, .... 771,

Can you sort them out now?

[russet]

Question 1: Find the smallest number which is divisible by 2,3,4,5 and 6 then take one away from it.

Question 2: Start by finding how many have only two sevens. If there were a 7 in the hundreds column and a 7 in the tens column there would be 9 other digits you could put in the units. Similarly if there were a 7 in the hundreds and a 7 in the units there would be 9 others for the tens column. If there were a 7 in the tens and units there would be eight others you could put in the hundreds. (you cannot start with zero). This only then leaves 777 to add for one more.

Question 3: Remember that if you have worked out how many ways there are of arranging 2,2,1 there will be the same number or ways of arranging 3,1,1.

Question 4. Imagine all the beads are lettered a to d. There are 4 possible ways of starting the necklace. Say you start with bead a. There are then 3 possible choices for the second bead; b,c or d. If you start with bead b, you still would have 3 choices left for the second bead. Same for starting with c or d. So for the first two beads there are 4*3 possibilities. Suppose you made up these 12 necklaces. For each necklace you would be left with 2 beads to choose from. This gives you 12*2 possibilities.
Now if a and b were the same colour, you would have a necklace running a,b,c,d which would look identical to b,a,c,d. In fact, out of the 24 necklaces, you would be able to put them into 12 pairs that looked the same, so you would only have 12 different colour combinations.

As a matter of interest you can write the sum above as 4!/2! where 4! means 4*3*2*1. It is read as "four factorial", and I think it is beyond GCSE.

[Guest55]

I really don't think it's helpful to include factorials ...

[Edited by moderator]

[NanoNano]

55, I think maybe the reason why Russet posted is because he likes to solve problems! I had to seriously stop myself from posting answers or suggestions myself as I love a puzzle/challenge! Instead, having noticed the protocol on this site, I just quietly worked them out in my notebook

[russet]

The OP was asking for the fastest way. I only mentioned factorials after showing how to answer the question without needing this knowledge.

[Edited by moderator]

[Martin_Procter]

I have been holding my peace for a while but I feel I have to post now.

The purpose of this forum is to share information and help people. There is often more than one way to solve a problem, the more information (within reason) the better.

[Edited by moderator]


NanoNano – don’t be put off. I, for one, would like to see your contribution.

[NanoNano]

Thanks Martin; not sure mine will be much of a contribution though! I love puzzles like Q1 (probably why I passed the 11+ back in the days when life in Bucks was simple!).

I fully understand G55's solution (so simply put, very neat and precise) but I approached it another way, and I will explain in case anyone else finds it easier to work in the same way (not sure that will be the case, but it is good to know different ways in order to find one that fits). Anyway, I just listed the 6 times table in one column (because that seemed the most challenging number), added 5 in the next column (6 remainder 5) and from there you can more of less eliminate most of the multiples as you go back through the rules of the question.

I know that sounds long-winded but really it isn't and is just methodical (though not as formulaic as G55's answer!).

[nikku]

Guest55, Russet, NanoNano
Thank you so much for replying to my post, I will work through the questions with your suggestions.