Limas soal Grup D

Suppose 𝑑 and 𝑗 are natural numbers so that 𝑗 < 2023

and

(𝑑^2 + 𝑑 + 1)(𝑑^2 + 3𝑑 + 3) = 𝑗^2 + 𝑗 + 1.

Determine the largest value of 𝑗.

Mathematical Equivalence

ON

Answer explanation

We can factor the left-hand side of the equation as [(d^2 + d + 1)(d^2 + 3d + 3) = (d + 1)(d + 1)(d + 3)(d + 1) = (d + 1)^2(d + 3).]Therefore, we have [(d + 1)^2(d + 3) = j^2 + j + 1.]Since j<2023, we have j2+j+1<20232+2023+1=409206. Therefore, d+1<409206​<202. Since d is a natural number, d≤201. Therefore, the largest possible value of j is 201​.

All the positive integers are arranged in a triangular array as shown below. Find the different of the number of the column and the number of the row where 2002 is put.

1 3 6 10 15 ...

2 5 9 14 ...

4 8 13 ...

7 12 ...

11 ...

Mathematical Equivalence

ON

Answer explanation

The difference between the number of the column and the number of the row where 2002 is put is 12​.

The numbers in the triangular array are the sum of the numbers in the two rows above it. For example, the number 15 is the sum of the numbers 3 and 12. The number 2002 is the sum of the numbers 1997 and 5. The number 1997 is the sum of the numbers 1993 and 4, and the number 1993 is the sum of the numbers 1989 and 4. Continuing in this way, we can see that the numbers 1989, 1985, ..., 5 are all in the first row of the triangular array. Therefore, the number 2002 is in the $\boxed{12}$th row of the triangular array.

Find the number of natural numbers 𝑛 such that 2^𝑛 + 5^𝑛 − 65 is a perfect square number.

Mathematical Equivalence

ON

Answer explanation

Let m2=2n+5n−65. For n=1, we have m2=9, which is a perfect square. For n=2, we have m2=576=242, which is also a perfect square. However, for n≥3, we have [m^2 = 2^n + 5^n - 65 = 2^n + 5^n - (5^2 - 2^2) = (2^n + 5^n) - (5 - 2)(5 + 2) = (2^n + 5^n) - 21.]Since 2n and 5n are always odd, 2n+5n is always even. Therefore, m2 is always even, and cannot be a perfect square. Hence, the only natural numbers n such that 2n+5n−65 is a perfect square are n=1 and n=2. Therefore, the answer is 2​.

If a random divisor of 10^10 is chosen, what is the probability that it divides at least one of 1000, 1250, or 1280?

Mathematical Equivalence

ON

Answer explanation

Let's first count the number of divisors of 10^10. A divisor of 10^10 can be written in the form 2^a*5^b, where 0 <= a <= 10 and 0 <= b <= 10. There are 11 choices for a and 11 choices for b, so there are 121 divisors of 10^10.

Now, let's count the number of divisors of 1000, 1250, and 1280. 1000 = 2^35^3, 1250 = 2^35^4, and 1280 = 2^7*5. The number of divisors of 1000 is (3+1)(3+1) = 16, the number of divisors of 1250 is (3+1)(4+1) = 25, and the number of divisors of 1280 is (7+1)(1+1) = 16. The number of divisors of 1000, 1250, and 1280 is 16+25+16 = 57.

The probability that a random divisor of 10^10 divides at least one of 1000, 1250, or 1280 is the number of divisors that divide at least one of 1000, 1250, or 1280 divided by the total number of divisors of 10^10. This is 57/121 = 19/47​​.

The point 𝑃 inside Δ𝐴𝐵𝐶 such that ∠𝐴𝐵𝑃 = 20° and ∠𝐴𝐶𝑃 = 15°. Find ∠𝐵𝑃𝐶 − ∠𝐵𝐴𝐶 in degrees

Mathematical Equivalence

ON

Answer explanation

Since ∠ABP+∠ABC+∠BCP=180∘ and ∠ABP=20∘, we have ∠ABC+∠BCP=160∘. Similarly, since ∠ACP+∠ABC+∠BCP=180∘ and ∠ACP=15∘, we have ∠BCP=105∘.

Therefore, ∠BPC=180∘−∠BCP=75∘​.

Suppose that 𝐴𝐵𝐶𝐷 is a trapezoid in which 𝐴𝐷 ∥ 𝐵𝐶. Given 𝐴𝐶 ⊥ 𝐶𝐷, 𝐴𝐶 bisects angle ∠𝐵𝐴𝐷 and area [𝐴𝐵𝐶𝐷] = 42, then compute area [𝐴𝐶𝐷]

Mathematical Equivalence

ON

Answer explanation

Let E be the foot of the perpendicular from A to BC. Then AB=AE+EB and CD=CE+ED. Since AC bisects ∠BAD, we have AE=CE and EB=ED. Therefore, AB=CD.

Let x=AE=CE.

Then the area of trapezoid ABCD is [\frac{1}{2}(AB + CD)(x) = \frac{1}{2}(2x)(x) = x^2.]

Since the area of trapezoid ABCD is 42, we have x2=42, so x=42​. Therefore, the area of triangle ACD is 21​(x)(x)=21​.

Let 𝑥, 𝑦, 𝑧 be the real numbers satisfying 2𝑥𝑦 − 𝑧^2 ≥ 1 and 𝑧 − |𝑥 + 𝑦| ≥ −1.

Calculate 20𝑥^2 + 2𝑦^2 + 3𝑧^2

Mathematical Equivalence

ON

Answer explanation

Squaring the first inequality, we get [4x^2y^2 - 4x^2z^2 + y^2z^2 \ge 1.]

Adding 2x2z2+2y2z2 to both sides, we get [4x^2y^2 + 6x^2z^2 + y^2z^2 \ge 1 + 2x^2z^2 + 2y^2z^2 = 3z^2.]

Since \ 4x2y2+6x2z2+y2z2≤4x2y2+2x2z2+2y2z2, we have [4x^2y^2 + 2x^2z^2 + 2y^2z^2 \le 3z^2.]

Multiplying both sides by 5, we get [20x^2 + 2x^2z^2 + 2y^2z^2 \le 15z^2.]

Combining this with the second inequality, we get [20x^2 + 2y^2 + 15z^2 \le 15z^2 + 3(-1) = 12z^2.]

Therefore, [20x^2 + 2y^2 + 3z^2 \le \boxed{12}.]

Let 𝑎 be a real number satisfying the equation

(((1+𝑎)^2)/(1+𝑎^2) = (20/23)).

Then ((1+𝑎)^3)/(1+𝑎^3) = (𝑑/𝑗)),

where 𝑑 and 𝑗 are positive coprime integers.

Find 𝑑 + 𝑗.

Mathematical Equivalence

ON

Answer explanation

We can solve the given equation for a to get [a = \frac{-1 \pm \sqrt{5}}{4}.]Since a is real, we must have a=4−1+5​​. Substituting this value into the second equation, we get [\frac{\left( \frac{-1 + \sqrt{5}}{4} \right)^3}{\left( \frac{-1 + \sqrt{5}}{4} \right)^3} = \frac{d}{j}.]This simplifies to [\frac{121 - 35 \sqrt{5}}{64} = \frac{d}{j}.]Therefore, d+j=185​.