agentofchange1:


oh..i see , guess my boss above has done(is doing) justice already ..

U can assist him naaaa.

Though i av the solutions with me already, jst wanna c if dia ar beta approach

ladokuntlad:
House i av a cowbell question to give

x+y+z=1
x²+y²+z²=35
x³+y³+z³=97

Show workings if x=-1 y=-3 and z=-3

partial solution

it can be established that

(x+y+z)^2 = x^2+y^2 +z^2 -2( xy +xz+ yz)

also

(x+y+z)[x^2 +y^2 +z^2 -(xy +xz +yz) ] = x^3 +y^3 +z^3 -3xyz

so from there , make necessary substitutions den solve to get final soln .


kinda lazy to solve 2day .. hope u get ?

All the 7-digit numbers containing each of the digits 1,2,3,4, 5,6, 7 exactly once, and not divisible by 5, are arranged in the increasing order. Find the 2000-th number in this list.

Show detailed working.

ladokuntlad:

U can assist him naaaa.

Though i av the solutions with me already, jst wanna c if dia ar beta approach

ok , what are the unsolved ones left ? , lets c if i can forward it(them) to ifa......

agentofchange1:


partial solution

it can be established that

(x+y+z)^2 = x^2+y^2 +z^2 -2( xy +xz+ yz)

also

(x+y+z)[x^2 +y^2 +z^2 -(xy +xz +yz) ] = x^3 +y^3 +z^3 -3xyz

so from there , make necessary substitutions den solve to get final soln .


kinda lazy to solve 2day .. hope u get ?

I dont want a lazy solution

Complete solution abeg

agentofchange1:


ok , what are the unsolved ones left ? , lets c if i can forward it(them) to ifa......

All av been solved
Jst hoping to get a beta solution

Given a positive integer k and other two integers b> w > 1. There are two strings of pearls, a string of b black pearls and a string of w white pearls. The length of a string is the number of pearls on it.
One cuts these strings in some steps by the following rules. In each step:

(i) The strings are ordered by their lengths in a non-increasing order. If there are some strings of equal lengths, then the white ones precede the black ones. Then k first ones (if they consist of more than one pearl) are chosen; if there are less than k strings longer than 1, then one chooses all of them.

(ii) Next, one cuts each chosen string into two parts differing in length by at most one.
(For instance, if there are strings of 5, 4, 4, 2 black pearls, strings of 8, 4, 3 white pearls and k 4, then the strings of 8 white, 5 black, 4 white and 4 black pearls are cut into the parts (4,4), (3,2), (2,2) and (2,2), respectively.)
The process stops immediately after the step when a first isolated white pearl appears.


Prove that at this stage, there will still exist a string of at least two black pearls.

Karmanaut:
As an alternative to the graphical method for question 4.
We're trying to see how many integer solutions the equation (x²-5×+5)^x[sup]2-11×+30[/sup] = 1 has.
Consider the equation a^b = 1.
The roots of this equation are a=1, b=0, a=1, b=1 also satisfies it.
Or if a=-1, and b is an even number.
Now let the base x² -5× +5 = a, and the power x²-11×+30 = b.

Case 1: a= 1
x² -5× + 5 = 1
The solutions are: x= 4, x=1.
We have gotten 2 valid answers.

Next:
b=0
x²-11×+30=0
The roots are: x=6, x=5.
Two more valid solutions.

Case II: b=1
ײ -11× + 30 = 1
Here we get two floating point solutions: x=6.61... and x= 4.38...
We discard them as they're not valid.

Case III: a=-1
ײ-5×+5=-1
The roots are: x=3, x=2.
We have two more valid solutions totalling 6.

Next:
b is even, let's start with 2.
x² -11×+30 = 2
The solutions are: x=7, x= 4
7 is not a valid solution as it is odd.
Only 4 satisfies the equation and we've gotten it before.

So it has 6 integer solutions: x=1,2,3,4,5,6.

Nbanu error at case 3 though answer is right
Na case 111 part b is even spoil d tin
Since at a=-1 gives soultion at x=3 or x=2
All we need todo is insert dose to values to b and see if its even
Case 1: x=2 insert inside x2-11x+30 gives 12=even valid
Case 2: x=3 insert also gives -60=even valid
Thus we av 6 different integers= 1,2,3,4,5,6

shaboti:
thank you

You welcome.

Mathschic never run away

Eat some cakes before leaving

Scroll to previous page and take a bite

ladokuntlad:
CONTINUATION OF THE AFOREMENTIONED QUESTIONS

Could you please type out the questions for some of us using local phone, thank you.

ladokuntlad:
Guys i av some maths Olympiad questions for u to solve

A three-digit number N has first digit a (≠0), second digit b and third digit c. N = b(10c + b) where b and (10c +b) are primes. Find N.

N has three digits in it, so N = 100a + 10b + c

N also is, N=b(10c+b), and b & 10c+b are primes

This means b is a single digit prime, while 10c + b is a double digit prime.

Since b is a single digit prime, b can be any of 2, 3, 5 & 7 which are the single digit primes possible.
Now, since 10c+b is also a prime, that means it is a double digit prime which ends with the digit b. The digit c multiplied by 10 will produce a number that has 0 in the second digit. This number will add to digit b and produce a prime. Thus, for 10c+b to be a prime, b cannot be 2 or 5, which don't end a double-digit prime.

Thus b can only be 3 or 7.

Now, 10c+b is also prime.
And N = b(10c+b); 10c+b = cb, with c in the tenth and b in the unit. Both are single digits.
N = cbxb. Since b is multiplying itself to produce a value in the unit, and b can either be 3 or 7; and also 3.3=9, 7.7=49, we have a unit value of 9. Hence c=9, since c is the unit on number N.

Thus N = b(10c+b) = b(90+b).

Now b can either be 3 or 7. Let's try out 3, N = 3(90+3) = 279; here b = 7, which is inconsistent with our assumption of b=3.

Let's try b=7
N = 7(90+7)= 679. Here b =7, which is consistent with our assumption.

This means N = 679

ladokuntlad:
Mathschic never run away

Eat some cakes before leaving

Scroll to previous page and take a bite

Don't worry, I'm here. Interesting things going on here; good work guys.

MathsChic:

A three-digit number N has first digit a (≠0), second digit b and third digit c. N = b(10c + b) where b and (10c +b) are primes. Find N.

N has three digits in it, so N = 100a + 10b + c

N also is, N=b(10c+b), and b & 10c+b are primes

This means b is a single digit prime, while 10c + b is a double digit prime.

Since b is a single digit prime, b can be any of 2, 3, 5 & 7 which are the single digit primes possible.
Now, since 10c+b is also a prime, that means it is a double digit prime which ends with the digit b. The digit c multiplied by 10 will produce a number that has 0 in the second digit. This number will add to digit b and produce a prime. Thus, for 10c+b to be a prime, b cannot be 2 or 5, which don't end a double-digit prime.

Thus b can only be 3 or 7.

Now, 10c+b is also prime.
And N = b(10c+b); 10c+b = cb, with c in the tenth and b in the unit. Both are single digits.
N = cbxb. Since b is multiplying itself to produce a value in the unit, and b can either be 3 or 7; and also 3.3=9, 7.7=49, we have a unit value of 9. Hence c=9, since c is the unit on number N.

Thus N = b(10c+b) = b(90+b).

Now b can either be 3 or 7. Let's try out 3, N = 3(90+3) = 279; here b = 7, which is inconsistent with our assumption of b=3.

Let's try b=7
N = 7(90+7)= 679. Here b =7, which is consistent with our assumption.

This means N = 679

419 method
Bt u try ooo no b small
U got d final answer

ladokuntlad:
Nbanu error at case 3 though answer is right Na case 111 part b is even spoil d tin Since at a=-1 gives soultion at x=3 or x=2 All we need todo is insert dose to values to b and see if its even Case 1: x=2 insert inside x2-11x+30 gives 12=even valid Case 2: x=3 insert also gives -60=even valid Thus we av 6 different integers= 1,2,3,4,5,6

Noted. Cheers.

The answer to question 7 is 679.

Solution:
We're looking for 3 digit N such that b and 10c+b are prime.
Since b is prime that means it is relatively prime to the other integers, such that gcd(a,b) = gcd(b,c) = 1.
Since 10c + b is prime we apply Dirichlet's theorem that if gcd(a,b) =1 then ax+b for some x is infinitely prime.
I'm not going to prove Dirichlet's theorem.

Applying the theorem to our problem we have gcd(b,c) = 1 (We're working with only b and c since we weren't given any information about a other than it is not equal to zero)

If b is a prime integer less than 10 then the set of possible values of b is given as {2, 3, 5, 7}
Therefore b can be any if those values.

Case 1:
B= 2.
The numbers less than 10 relatively prime to 2 are: 1, 3, 5, 7, 9
So the possible value of c is in the set {1, 3, 5, 7, 9}
Then we use trial and error to check if they satisfy the condition that 10c + b is prime.
We actually do not need trial and error in this case because setting b as 2, the condition becomes: 10c + 2 is prime, which is impossible for all values of c because 10c+2 can be factorised as 2(5c+1) in which for all values of c, the answer is clearly a multiple of 2 and is therefore composite.

Case II:
B=3
If b=3, then the numbers less than 10 relatively prime are: 1,2,4,5,7,8
So the possible value of c when b=3 is a member of the set {1, 2, 4, 5, 7, 8}
Here we use trial and error to check if 10*c + b is prime because 10*c +3 cannot be factorised.
If c=1, we have 10*1 + 3 = 13 which is prime.
If c=2, we have 10*2 + 3 = 23 which is prime.
If c=4, we have 10*4 + 3 =43 which is prime.
If c=5, we have 10*5 + 3 =53 which is prime.
If c=7, we have 10*7 + 3 = 73 which is prime.
If c=8 we have 10*8 + 3 = 83 which is prime.

Next we test the values one after the other to see if they satisfy the other condition N is 3 digit with N = b(10c+b) with b and c retaining their positions in N.
Clearly 13 and 23 are out of it because they don't give a 3 digit number.
Also b cannot be = 3 because all values of 10c + b end in 3, and b=3* (10c+b) will give answers that end in 9, and 9 is not part of the possible values of c because it is not relatively prime to 3.

Case III:
b = 5.
Clearly this won't work either because 10c+ b with b=5 is a composite number, with 5 as a factor of it.
10c+ 5 = 5(2c +1)

Case IV:
B= 7.
10c+ b cannot be factored by 7 or 10 so 7 is a valid candidate.
The numbers less than 10 relatively prime to 7 are: 1,2,3,4,5,6,8,9
Therefore the possible value of c can be found in the set: {1, 2, 3, 4, 5, 6, 8, 9}
Now we test via trial and error if 10c+ b is prime.
If c=1, 10*1 + 7 =17 which is prime
If c=2, 10*2 + 7 = 27 which is composite as it is divisible by 9 and 3.
If c= 3, 10*3 + 7 =37 which is prime.
If c=4, 10*4 + 7 = 47 which is prime.
If c=5, 10*5 + 7 =57 which is composite as it has factors 3 and 17.
If c= 6, 10*6 + 7 = 67 which is prime.
If c= 8, 10*8 + 7 =87 which is composite as it has factors 3 and 29.
If c= 9, 10* 9 + 7 = 97 which is prime.

So the possible values of 10c+b when b = 7 are:
17, 37, 47, 67, 97.
So we test b(10c+b) , 17 is not a candidate as 7*17 gives a 2 digits number instead of 3.
For an answer to be valid, b*(10c+b) has to be 3 digits with b and c retaining their positions and values.
37 * 7 = 259
47* 7 = 329
67* 7 = 469
97* 7 = 679.

We have a winner!
259, 329, and 469 are not valid because b and c do not retain their values an positions, moreover in 469, 6 and 9 are not relatively prime.

679 is clearly the answer.
A= 6, b = 7, c = 9
It satisfies all conditions :
B= 7 is prime, and 10c+ b = 97 is also prime, b(10c+b) is 3 digit and b and c retain their values and positions.
Q.E.D

Karmanaut:
The answer to question 7 is 679.

Solution:
We're looking for 3 digit N such that b and 10c+b are prime.
Since b is prime that means it is relatively prime to the other integers, such that gcd(a,b) = gcd(b,c) = 1.
Since 10c + b is prime we apply Dirichlet's theorem that if gcd(a,b) =1 then ax+b for some x is infinitely prime.
I'm not going to prove Dirichlet's theorem.

Applying the theorem to our problem we have gcd(b,c) = 1 (We're working with only b and c since we weren't given any information about a other than it is not equal to zero)

If b is a prime integer less than 10 then the set of possible values of b is given as {2, 3, 5, 7}
Therefore b can be any if those values.

Case 1:
B= 2.
The numbers less than 10 relatively prime to 2 are: 1, 3, 5, 7, 9
So the possible value of c is in the set {1, 3, 5, 7, 9}
Then we use trial and error to check if they satisfy the condition that 10c + b is prime.
We actually do not need trial and error in this case because setting b as 2, the condition becomes: 10c + 2 is prime, which is impossible for all values of c because 10c+2 can be factorised as 2(5c+1) in which for all values of c, the answer is clearly a multiple of 2 and is therefore composite.

Case II:
B=3
If b=3, then the numbers less than 10 relatively prime are: 1,2,4,5,7,8
So the possible value of c when b=3 is a member of the set {1, 2, 4, 5, 7, 8}
Here we use trial and error to check if 10*c + b is prime because 10*c +3 cannot be factorised.
If c=1, we have 10*1 + 3 = 13 which is prime.
If c=2, we have 10*2 + 3 = 23 which is prime.
If c=4, we have 10*4 + 3 =43 which is prime.
If c=5, we have 10*5 + 3 =53 which is prime.
If c=7, we have 10*7 + 3 = 73 which is prime.
If c=8 we have 10*8 + 3 = 83 which is prime.

Next we test the values one after the other to see if they satisfy the other condition N is 3 digit with N = b(10c+b) with b and c retaining their positions in N.
Clearly 13 and 23 are out of it because they don't give a 3 digit number.
Also b cannot be = 3 because all values of 10c + b end in 3, and b=3* (10c+b) will give answers that end in 9, and 9 is not part of the possible values of c because it is not relatively prime to 3.

Case III:
b = 5.
Clearly this won't work either because 10c+ b with b=5 is a composite number, with 5 as a factor of it.
10c+ 5 = 5(2c +1)

Case IV:
B= 7.
10c+ b cannot be factored by 7 or 10 so 7 is a valid candidate.
The numbers less than 10 relatively prime to 7 are: 1,2,3,4,5,6,8,9
Therefore the possible value of c can be found in the set: {1, 2, 3, 4, 5, 6, 8, 9}
Now we test via trial and error if 10c+ b is prime.
If c=1, 10*1 + 7 =17 which is prime
If c=2, 10*2 + 7 = 27 which is composite as it is divisible by 9 and 3.
If c= 3, 10*3 + 7 =37 which is prime.
If c=4, 10*4 + 7 = 47 which is prime.
If c=5, 10*5 + 7 =57 which is composite as it has factors 3 and 17.
If c= 6, 10*6 + 7 = 67 which is prime.
If c= 8, 10*8 + 7 =87 which is composite as it has factors 3 and 29.
If c= 9, 10* 9 + 7 = 97 which is prime.

So the possible values of 10c+b when b = 7 are:
17, 37, 47, 67, 97.
So we test b(10c+b) , 17 is not a candidate as 7*17 gives a 2 digits number instead of 3.
For an answer to be valid, b*(10c+b) has to be 3 digits with b and c retaining their positions and values.
37 * 7 = 259
47* 7 = 329
67* 7 = 469
97* 7 = 679.

We have a winner!
259, 329, and 469 are not valid because b and c do not retain their values an positions, moreover in 469, 6 and 9 are not relatively prime.

679 is clearly the answer.
A= 6, b = 7, c = 9
It satisfies all conditions :
B= 7 is prime, and 10c+ b = 97 is also prime, b(10c+b) is 3 digit and b and c retain their values and positions.
Q.E.D

Mathschic got it first.
Bt wait oo
D whole tin naa 4hrs and u guys don spend 8hrs already

ladokuntlad:
Mathschic got it first.

I saw her post after I posted mine.

Bt wait oo D whole tin naa 4hrs and u guys don spend 8hrs already

Some of us have lives you know.

thankyouJesus:
Could you please type out the questions for some of us using local phone, thank you.

Truth b told, its lenthy
And some maths notations cant b type here

Karmanaut:

I saw her post after I posted mine.

Some of us have lives you know.

Dats true shaaa

ladokuntlad:
Truth b told, its lenthy And some maths notations cant b type here

okay sir

Try...


for. 0<x<2π

Karmanaut:

Some of us have lives you know.

Lol. True

ladokuntlad:

419 method
Bt u try ooo no b small
U got d final answer

Haha, 419? I thought I was clear and logical enough

MathsChic:
Haha, 419? I thought I was clear and logical enough

this chick, u don com again

LORDDICE:



this chick, u don com again

Lorddice you don come too.

MathsChic:

Haha, 419? I thought I was clear and logical enough

don't mind me jare.
Am impressed..
Try attempting 1,2,3 of latest screenshot

Madmathecian:
Given a positive integer k and other two integers b> w > 1. There are two strings of pearls, a string of b black pearls and a string of w white pearls. The length of a string is the number of pearls on it.
One cuts these strings in some steps by the following rules. In each step:

(i) The strings are ordered by their lengths in a non-increasing order. If there are some strings of equal lengths, then the white ones precede the black ones. Then k first ones (if they consist of more than one pearl) are chosen; if there are less than k strings longer than 1, then one chooses all of them.

(ii) Next, one cuts each chosen string into two parts differing in length by at most one.
(For instance, if there are strings of 5, 4, 4, 2 black pearls, strings of 8, 4, 3 white pearls and k 4, then the strings of 8 white, 5 black, 4 white and 4 black pearls are cut into the parts (4,4), (3,2), (2,2) and (2,2), respectively.)
The process stops immediately after the step when a first isolated white pearl appears.


Prove that at this stage, there will still exist a string of at least two black pearls.

Nobody is going for my question?

Madmathecian:



Nobody is going for my question?

am on it

MathsChic:

Lorddice you don come too.

hehehe, I never come o.. I was having my shower, I just sited u and decided to say hi....

LORDDICE:




hehehe, I never come o.. I was having my shower, I just sited u and decided to say hi....

Oh cool. Sayin hi here too.

MathsChic:
Oh cool. Sayin hi here too.

goodnyt o. let me sleep. cos I know say u no dey sleep