Combinatoric discussion

Problem 4:

What is the maximum number of Bishops you can place on an 8×8 chessboard? without any bishops attacking each other.

And identify example of those bishops coordinates.

You can place 14 bishops without any attacking one another. One example is 8 bishops all on the 1st row, and 6 on the 8th row. However, none of the bishops on the 8th row can be at a8 or h8

That's right. To be clear, I've proved the following.

The position of the 14 bishops are represented by yellow highlight.

Notice that two bishops cannot be placed on the same numbered position, otherwise they would be on the same diagonal.

Therefore, if there are 15 bishops, at least 2 bishops must be placed on the same numbered position. Which is impossible.

Problem 5:

Level : difficult (edited)

Let a,b ∈ {1,2,3,...,100} find the number of possible values ​​of aᵇ.

Notes: a and b may or may not be equal.

This is a problem that I set up, and just successfully resolved. It's harder than I thought.

You can request a hint or solution from me.

First of all, there are 100X100=10,000 possible non-distinct values of a^b

The only way an a^b could be equal to another a^b is if the a is squared, cubed, quarted, quinted, etc…

Case 1: 1^x where x is an element of the set [1, 2, 3…100] is all equal to 100. Take out 99 possible values.

Case 2:

2^2 takes out all even values of b, so subtract 50. 2^3 takes out all odd multiples of 3 less than 100, which is 16. 2^4 is a repeat of 2^2, so it is irrelevant. 2^5 takes out 25, 35, 55, 65, 85, 95, which is 6 values. 2^6 will repeat 2^3 and 2^2.

Case 3:

3^2 takes out all even values of b, which is 50. 3^3 take out all odd multiples of 3 less than one hundred, which is 16. 3^4 is irrelevant.

Case 4: 5^2 takes out all even values of b, which is 50.

Case 5: 6^2 takes out all even values of b, which is 50.

Case 6: 7^2 takes out all even values of b, which is 50.

Case 8: 10^2 takes out all even values of b, which is 50.

If we add all the same values together, we get 99+50+16+6+50+16+50+50+50+50=437.

There are 437 repeating values of a^b where a and b are elements of the set [1, 2, 3,…, 100]
Subtract 437 from 10,000 and you will have your answer.

10,000-437=9563.

There are 9,563 distinct values for a^b where a and b are elements of the set [1, 2, 3,…, 100].

I’m not sure if that is right

It’d be a good question on the AIME

It seem that the answer I got was 9,414 distinct value, I will post a solution soon.

I separately the base of a power that is not a perfect power n

e.g. 4¹⁰⁰=16⁵⁰ and equals 2²⁰⁰, but this doesn't have an exponent in the scope. So I'll consider the answer in the form 2ˣ, 3ˣ, ... without limiting x<100.

Solution coming soon...

Oh I forgot to add in the other composite numbers thinking they were irrelevant. You’re correct, and I’ll add them soon to my solution

Sorry I miscalculated by 1 point. The answer should be 9,418. I will check for sure. Before posting how to do it in detail.

This is how I started.

I give aᵇ = uˣ when u cannot be written in exponent form with bases and exponents being integers greater than 1 and x is an integer.

aᵇ = 1ˣ -> There is only 1 value

aᵇ = 2ˣ -> Since 2,4,8,16,32 and 64 all have 2 in their power base. So I convert all bases to 2 with identity (aᵐ)ⁿ = aᵐⁿ.

When a=4 ; aᵇ yields [4¹,4²,4³,...,4¹⁰⁰] = [2²,2⁴,2⁶,...2²⁰⁰]. Similarly, aᵇ yields [2³,2⁶,2⁹,.. .2³⁰⁰], [2⁴,2⁸,2¹²,...2⁴⁰⁰], [2⁵,2¹⁰,2¹⁵,...2⁵⁰⁰], [2⁶,2¹²,2¹⁸,...,2⁶⁰⁰] when a=8,16,32 and 64 respectively.

So the task now is to find all the x which 2ˣ that given by aᵇ now...

This is my solution (answer is 9,271)

I give aᵇ = uˣ when u cannot be written in exponent form with bases and exponents being integers greater than 1 and x is an integer.

aᵇ = 1ˣ -> There is only 1 value

aᵇ = 2ˣ -> Since 2,4,8,16,32 and 64 all have 2 in their power base. So I convert all bases to 2 with identity (aᵐ)ⁿ = aᵐⁿ.

When a=4 ; aᵇ yields [4¹,4²,4³,...,4¹⁰⁰] = [2²,2⁴,2⁶,...,2²⁰⁰]. Similarly, aᵇ yields [2³,2⁶,2⁹,...,2³⁰⁰], [2⁴,2⁸,2¹²,...,2⁴⁰⁰], [2⁵,2¹⁰,2¹⁵,...,2⁵⁰⁰], [2⁶,2¹²,2¹⁸,...,2⁶⁰⁰] when a=8,16,32 and 64 respectively.

So the task now is to find all the x which 2ˣ that given by aᵇ now. Let aₖ be the set of x in each interval that is divisible by k.

1 ≤ x ≤ 100 ; x is all possible. That is 100 difference value.

101 ≤ x ≤ 200 ; The number of possible x is n(a₂)+n(a₃)+n(a₅)-n(a₆)-n(a₁₀)-n(a₁₅)+n(a₃₀) = 50+33+20-17-10-7+3 = 72

201 ≤ x ≤ 300 ; The number of possible x is n(a₃)+n(a₄)+n(a₅)-n(a₁₂)-n(a₁₅)-n(a₂₀)+n(a₆₀) = 34+25+20-9-7-5+2 = 60

301 ≤ x ≤ 400 ; The number of possible x is n(a₄)+n(a₅)+n(a₆)-n(a₁₂)-n(a₂₀)-n(a₃₀)+n(a₆₀) = 25+20+16-8-5-3+1 = 46

401 ≤ x ≤ 500 ; The number of possible x is n(a₅)+n(a₆)-n(a₃₀) = 20+17-3 = 34

501 ≤ x ≤ 600 ; The number of possible x is n{504,510,516,...,600} = 17

Sum of all case 2ˣ = 100+72+60+46+34+17 = 329

aᵇ = 3ˣ -> Since 3,9,27 and 81 all have 3 in their power base. in the same way as 2ˣ we will get the value [3²,3⁴,3⁶,...,3²⁰⁰], [3³,3⁶,3⁹,...,3³⁰⁰], [3⁴,3⁸,3¹²,...,3⁴⁰⁰] when a=9,27 and 81 respectively.

1 ≤ x ≤ 100 ; x is all possible. That is 100 difference value.

101 ≤ x ≤ 200 ; The number of possible x is n(a₂)+n(a₃)-n(a₆) = 50+33-17 = 66

201 ≤ x ≤ 300 ; The number of possible x is n(a₃)+n(a₄)-n(a₁₂) = 34+25-9 = 50

301 ≤ x ≤ 400 ; The number of possible x is n{304,308,312,...,400} = 25

Sum of all case 3ˣ = 100+66+50+25 = 241

aᵇ = uˣ for u∈{5,6,7,10} ; Since u and u² all have u in their power base and u² ≤ 100. we will get the value u²,u⁴,u⁶,...,u²⁰⁰ when a=u²

1 ≤ x ≤ 100 ; x is all possible. That is 100 difference value.

101 ≤ x ≤ 200 ; The number of possible x is n{100,102,104,...,200} = 50

Sum of all case uˣ for u∈{5,6,7,10} = 4×(100+50) = 600

aᵇ = uˣ for u∈{11,12,13,...,100} - {16,25,27,32,36,49,64,81,100} which is a set with 90-9 = 81 elements. We find that every u, u², u³,...u¹⁰⁰ are different.

Since we subtract the set of exponential numbers from the previous case (that can be converted to base no more than 10). So all the different values ​​for this case are 81×100 = 8,100

SUM OF ALL CASE = 1+329+241+600+8,100 = 9,271

Problem 6:

How many different triangles are there? with an integer side length and a perimeter of 20?

The relationship between the sides of the triangle can be expressed by a+b>c, where c is the longest side. If the perimeter is 20, then all the possible cases (in (a,b,c)) are (2,9,9), (3,8,9), (4,7,9), (4,8,8), (5,6,9), (5,7,8), (6,6,8), and (6,7,7). There are, in total, 8 possible triangles.

If the product of 5 positive real numbers is 3125, what is the minimum value of their sum?

10,080 I think

If we form it by a ring with English men all sitting next to American men and American men all sitting next to English men, then we can separate it into two separate lines, both of which are seven long. This means we can change the order in each of those lines 7! times. However, I think this is still wrong. It’s just a guess.

2*7!=10,080

That makes more sense

THIS IS ABOUT ALGEBRA

The HM-GM-AM inequality states that "harmonic mean ≤ geometric mean ≤ arithmetic mean".

For a₁,a₂,...,aₙ to be any positive real number, we get.

n/[(1/a₁)+(1/a₂)+...+(1/aₙ)] ≤ root_n(a₁a₂...aₙ) ≤ (a₁+a₂+...+aₙ)/n

The inequality will be an equation when a₁=a₂=...=aₙ

Let a,b,c,d,e ∈ R+ that abcde = 3125

By the AM-GM inequaility ; (a+b+c+d+e)/5 ≥ (abcde)^(1/5)

(a+b+c+d+e)/5 ≥ 3125^(1/5) = 5

a+b+c+d+e ≥ 25

Therefore, the minimum value of a+b+c+d+e is 25, where a=b=c=d=e=5.

Yes. we can set the first position for one person on the circle because rotation does not change the arrangement order on the circle.

So the remaining 6 people are viewed as a permutate on a straight line which is possible in 6! = 720 ways.