Maths- Gym for Brainiacs

ohh i see... made it clearer now

halamadrid2 wrote:ohh i see... made it clearer now

I-no is on 22th minute and still didn't solve it I solved it in a strange way, maybe there're better ways
Whoever manages to solve the problem, hats off to him/her

ok ive gotten as close as i could get lol

so if he was supposed to get 34.69 back he instead got 69.38 when he got home he had 69.33 am i on the right track

halamadrid2 wrote:ok ive gotten as close as i could get lol

so if he was supposed to get 34.69 back he instead got 69.33 when he got home he had 69.38 which is double the amount he had to get



so initial value is 100-34.69=65.31

He lost 5 cents on the way home. In your version, he gets 5 cents more magically :lol!:
Also, in your version, the initial bill would be 34.69

yh i corrected that... but am i getting closer

halamadrid2 wrote:yh i corrected that... but am i getting closer

Do you need hints? I didn't know how to solve the question either the first 3 minutes

Oh yeah, some goes to spanky, kiranr, lord hades etc. If you want hints, ask, if you don't want to solve it, say as well ( I can't read your minds)

i pass , i dont like solving these type of questions.

The book I copied the exercise from doesn't provide solutions If you find the right solution you could check yourself whether it's right or not So take my solution with caution, maybe there're better ways to solve it

Next question

topic: analysis
difficulty: very hard
requirements: junior high level

Hala2 is hungry. He opens his refrigirator and doesn't find anything With no choice left, he goes to the closest super market
He buys his stuff and pays with a 100€ bill. The cashier confuses euros and cents while giving the drawback. While Hala2 packs his stuff into bags, he loses 5 cents on the ground without noticing. At home, he notices that he's got twice the amount of drawback he should've gotten in the first place
How high was his super market bill initially?

First, I made a gross estimation. Say x is the unknown bill and y the drawback then it would look something like this:
100-x=y
If I substract an insignificant amount of 0.05€ from either x or y, their ratio will remain unchanged, roughly 1:2.
It means the bill is somewhat like: y=2x -> 100-x=2x -> x=100/3
The number I'm looking for is between 33.33 and 30.00 (gross estimation)

Next step, I gave an unknown varibale to every digit of the bill number:
ab.cd€ ( for example 19.01€ then a=1, b=9, c=0, d=1)

From the first step I know that a=3 and if I double a number which is between 33.33 and 30.00 then it will have a six in front therefore c=6.

Now, b and d are left. I know from the exercise that:
6d.3b - 0.05€ = 2* 3b.6d

If you take the digit I marked black times two there will be an extra decimal in place I marked red. So the relationship between b and d is:
d=2*b+1

From here on I tried to find a whole number( trial and error) which fits this relationship,
condition 1<d<9,
b could be 1,2,3 or 4.
I started off with 1 and it worked fine For b=1, d=2*1+1=3

Let's check

63.31-0.05=2* 31.63=63.26

Answer: the initial bill was 31.63€

Next question

topic: numerical theory
difficulty:medium
requirements: senior high school

The term n=10m+1 is given. n and m are whole and positive, m is an odd number.

Objective: show that the term n^5-51 is divisable by 100

Another question

topic: numerical theory
difficulty:medium
requirements: senior high school

Objective: show that if the checksum of a three digit number is divisable by 3 then the number itself is divisable by 3

Kamina wrote:Next question

topic: numerical theory
difficulty:medium
requirements: senior high school

The term n=10m+1 is given. n and m are whole and positive, m is an odd number.

Objective: show that the term n^5-51 is divisable by 100

ok... for that to be divisible by 100 it has to divide fully with 100s prime factors which are 2 and 5

now lets say n=10m+1

use the binomial expansion to show that:

(10m+1)^5-51=10m^5+5(10m^4)+10(10m^3)+10(10m^2)+5(10m)

to show that its divisible by 2:

let m=3

2430(even)+4050(even)+2700(even)+900(even)+150(even)=10230which is an even number hence its divisible by 2

and to show that its divisible by 5:

(n^5-51) and you take out a factor of 5 from the binomila expansion and you get:

5(2m^5+10m^4+20m^3+20m^2+10m)

this means that it is clearly divisible by 5 hence the term n^5-51 is divisible by 100

halamadrid2 wrote:

Kamina wrote:Next question

topic: numerical theory
difficulty:medium
requirements: senior high school

The term n=10m+1 is given. n and m are whole and positive, m is an odd number.

Objective: show that the term n^5-51 is divisable by 100

ok... for that to be divisible by 100 it has to divide fully with 100s prime factors which are 2 and 5

now lets say n=10m+1

use the binomial expansion to show that:

(10m+1)^5-51=10m^5+5(10m^4)+10(10m^3)+10(10m^2)+5(10m)

to show that its divisible by 2:

let m=3

2430(even)+4050(even)+2700(even)+900(even)+150(even)=10230which is an even number hence its divisible by 2

and to show that its divisible by 5:

(n^5-51) and you take out a factor of 5 from the binomila expansion and you get:

5(2m^5+10m^4+20m^3+20m^2+10m)

this means that it is clearly divisible by 5 hence the term n^5-51 is divisible by 100

First red mark: you only showed for m=3. What about m=11 ,101 or 529 ? .
Hint: you can substitute any odd number like this
m=2*a-1 where a is a whole positive number equal or greater than 1 .

Second red mark: you've proven that the term is divisable by 10 not by 100 . 100 is 2²*5² .

Overall, you're on the right track if you include the hint I gave you

for the first one i only used an example because its the same for all odd numbers.... which is why i showed "even" in brackets b/c thats the general rule for all odd numbers

arent the prime numbers 2,2,5,5??

does that mean i have to show it for 25 and 4 as well???

halamadrid2 wrote:for the first one i only used an example because its the same for all odd numbers.... which is why i showed "even" in brackets b/c thats the general rule for all odd numbers

arent the prime numbers 2,2,5,5??

does that mean i have to show it for 25 and 4 as well???

Divisable by 100 means a division with no rest. 100= (2*5)^2.
We aren't looking for a decimal number, we're looking for whole positive numbers. You have to show that the term is divisable by 100. Whether you write it as (2*5)², 2²*5² or 10² doesn't matter
Like I said use the hint I gave you, the answer will get apparent. The question isn't that hard

ok so basically since its divisible by both 2 and 5

you square root the expansions to prove that they are divisible by 2² and 5²

No, go directly with 100 or 20*5
If you do it properly, you'll need 3 lines at best to prove

ok....

i have proved it for 5 and 2

2430(even)+4050(even)+2700(even)+900(even)+150(even)=10230which is an even number hence its divisible by 2

5(2m^5+10m^4+20m^3+20m^2+10m) i.e. divisible by 5


now i have to prove it only for 10

from the binomial expanision:

(10m+1)^5-51=10m^5+5(10m^4)+10(10m^3)+10(10m^2)+5(10m)

you can see from the first term that its dvisible by 10 and so is the second one and the third one and also the last one

again you can take out a factor of ten to prove this

10(m^5+5m^4+10m^3+10m^2+5m)

all in all the this N^5-51 is therefore divisible by 100

halamadrid2 wrote:ok....

i have proved it for 5 and 2

2430(even)+4050(even)+2700(even)+900(even)+150(even)=10230which is an even number hence its divisible by 2

5(2m^5+10m^4+20m^3+20m^2+10m) i.e. divisible by 5


now i have to prove it only for 10

from the binomial expanision:

(10m+1)^5-51=10m^5+5(10m^4)+10(10m^3)+10(10m^2)+5(10m) -51?

you can see from the first term that its dvisible by 10 and so is the second one and the third one and also the last one

again you can take out a factor of ten to prove this

10(m^5+5m^4+10m^3+10m^2+5m)

all in all the this N^5-51 is therefore divisible by 100

Where did -51 get lost on the right side? (red) Also no, you can't just use a number for m. Substitute it by the hint I gave you Again, you have to prove it for all positive+whole unlimited amount of odd numbers which exist Just start off with m=2*a-1 and don't use your previous calculation at all and you'll see the light

ok let m=2a-1


10m^5+5(10m^4)+10(10m^3)+10(10m^2)+5(10m)=

=(10(2a-1)^5+ 50(2a-1)^4+100(2a-1)^3+11(2a-1)^2+50(2a-1))

for example let a=1

and you get 10+50+100+100+50=310 which is even and any even number is divisible by 2

and thats valid for any odd number

if this is wrong then i give up

Hala2
Look, first of all let's clear up what divisable means in maths. If 6 is divisable by 2 then there must exist a whole, positive number which times 2 is six:

6=2*a

a in this example is 3.

For example, I want to know whether the term

10a+20b with a,b natural numbers

is divisable by 5. How to show it? You have to show that

10a+20b=5*c

five times something whole and positive. You don't need to know the value of a or b at any point to prove this.

proof:

10a+20b=5*2a+5*4b=5*(2a+4b)=5*c

and we're done.


Now back to the above question:

What to prove?

You have to show:

n^5 - 51=100*c where c belongs to natural numbers

We know that n=10m+1 where m is odd, in other words m=2a-1. a belongs to natural numbers.
2a-1 represents any odd number, how? Simple:

2*1-1=1
2*2-1=3
2*3-1=5 and so on...

You don't need m's proper value at any point Now, to put you in the right way, substitute m through 2a-1 in n=10m+1 from the very start:

n=10m+1=10*(2a-1)+1=20a-10+1=20a-9

Now, to the proof, you start with:

n^5-51= (20a-9)^5 -51=...

Your objective is to transform the right side into 100 times something, 100*c for example where c is a natural number.


I'll write natural numbers instead of whole and positive numbers from now on

The set of natural numbers contains all whole, positive numbers. Example: 1, 2, 3, 4, 5, 6 and so on...

You're in no hurry to solve this but if you do you'll learn a whole lot about maths

(20a-9)^5-51=100c

20a^5(-9)^0+5x^4(-9)^1+10x^3(-9)^2+10x^2(-9)^3+5x(-9)^4+(-9)^5-51=

=3200000a^5-1440000a^4+6480000a^3-2916000a^2+656100a-59100=100c

= to show that its divisible by 100

you take out 100 as a common factor

100*32000a^5-100*14400a^4+100*64800a^3-100*296160a^2+100*6561a-100*591=

= 100(32000a^5-14400a^4+64800a^3-296160a^2+6561a-591)=100c

Good job hala2, you grasped the idea behind the proof

topic: numerical theory
difficulty:hard
requirements: senior high school

Objective: show that if the checksum of a three digit number is divisable by 3 then the number itself is divisable by 3

Here, a big hint. You already know what divisable means You could write a 3 digit number as xyz then it's devisable checksum would be:
x+y+z=3*a
where a belongs to the set of natural numbers.
It's a premise given by the exercise, you don't have to prove it but you need it for the proof Next, you could write a 3 digit number like this, for example:
931=9*100+3*10+1

Use your creativity and the knowledge about divisable numbers and you'll get the idea

yeah finally lol..... after being spoon fed haha... thanks tho

just going off to have my lunch.... will do it when i come back

Ask questions if you need help somewhere

399= 100(3)x+10(9)y+9z=3a=

=3*100x+3*10y+3*3z=3a

=3(100x+10y+9z)

halamadrid2 wrote:399= 100(3)x+10(9)y+9z=3a=

=3*100x+3*10y+3*3z=3a

=3(100x+10y+9z)

You proved for 399 only in your realm. How about all 3 digit numbers xyz?
All in all, the left side in what you did isn't equal to the right side unless x,y and z are always equal to 1
Start off with:
xyz=100x+10z+y=...

No progress peeps? Hala2, if you give up I'll write the answer