NMTC kaprekar ( sub junior level) 7th and 8th Aug 23 2025

In 15 minutes, that is 900 seconds, the two runners together cover $$(6 + 5) \times 900 = 9900$$ metres. Along a road of length 400 metres they come together once for every 400 metres of combined running. Since $$\frac{9900}{400} = 24.75$$, exactly 24 such meetings are completed, so they meet 24 times.

$$AB$$ is parallel to $$CD$$ and the two slanting lines carrying the arrow marks are parallel to each other. Using corresponding angles on the transversal through $$C$$ and $$E$$, the angle at $$C$$ above the line $$CD$$ is $$85^\circ$$, and alternate angles with $$AB$$ parallel to $$CD$$ give $$41^\circ$$ for the part of it lying above $$CD$$. Hence $$x = 180^\circ - 85^\circ - 41^\circ = 54^\circ$$.

$$\dfrac{1}{x+a} + \dfrac{1}{x+b} = \dfrac{1}{x+a+b} + \frac{1}{x}$$

$$\dfrac{2x+a+b}{(x+a)(x+b)}= \dfrac{2x+a+b}{x(x+a+b)} $$

So, for the equality to hold, we get $$2$$ cases here

CASE 1: $$2x+a+b=0$$

So, $$x=\dfrac{-(a+b)}{2}$$

CASE 2: $$2x+a+b\neq0$$

$$\dfrac{2x+a+b}{(x+a)(x+b)}= \dfrac{2x+a+b}{x(x+a+b)} $$

So, $$2x+a+b$$ can be cancelled out from both sides of the equality, as it's a non-zero number

$$\dfrac{1}{(x+a)(x+b)}= \dfrac{1}{x(x+a+b)} $$

Cross-multiplying, we get:

$$x(x+a+b)=(x+a)(x+b)$$

or, $$x^2+ax+bx=x^2+ax+bx+ab$$

$$ab=0$$

So, this means either $$a$$ or $$b$$ or both are zero.

That means the equality will be satisfied here at some particular value of $$a$$ and $$b$$ and not every value of $$a$$ and $$b$$

So, for any value of $$a$$ and $$b$$, the equality holds only if $$x=\dfrac{-(a+b)}{2}$$

Hence, Option D is correct.

The repeated side may be $$23$$ cm, giving perimeter $$23 + 23 + 17 = 63$$, or it may be $$17$$ cm, giving $$17 + 17 + 23 = 57$$. Both are genuine triangles because $$17 + 17 = 34 > 23$$. So the perimeter is $$63$$ or $$57$$.

The interior angles of a pentagon add up to $$540^\circ$$, so $$\angle A + \angle C + \angle D = 540^\circ - 90^\circ - 150^\circ = 300^\circ$$. Substituting $$\angle A = 180^\circ - \angle D$$ and $$\angle C = 180^\circ - \angle D$$ gives $$360^\circ - \angle D = 300^\circ$$, so the interior angle $$\angle D = 60^\circ$$. The external angle at $$D$$ is therefore $$180^\circ - 60^\circ = 120^\circ$$.

The unit digits of powers of 3 repeat in the block 3, 9, 7, 1 with period 4, and $$2025 = 4 \times 506 + 1$$, so $$3^{2025}$$ ends in 3. The unit digits of powers of 7 repeat as 7, 9, 3, 1, and 2024 is a multiple of 4, so $$7^{2024}$$ ends in 1. The product therefore ends in $$3 \times 1 = 3$$.

TO BE FILLED - printed options need checking. Since $$18900 = 2^2 \times 3^3 \times 5^2 \times 7$$, the exponents of 2, 5 and 7 must be raised to multiples of 3, which needs $$n = 2 \times 5 \times 7^2 = 490$$, a value that does not appear among the options printed for this problem.

Here $$a = 1.2c$$ and $$b = 1.5c$$, so $$\frac{a}{b} = \frac{1.2c}{1.5c} = \frac{4}{5} = 0.8$$. As a percentage this is 80 %.

Since $$\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} = 18$$, we get $$ab = \frac{2}{18} = \frac{1}{9}$$. Then $$a^3 + b^3 = (a+b)^3 - 3ab(a+b) = 8 - 3 \times \frac{1}{9} \times 2 = 8 - \frac{2}{3} = \frac{22}{3}$$. As $$\frac{22}{3} \approx 7.33$$, the value lies between 7 and 8.

Squaring both sides gives $$12 + \sqrt[3]{x} = \frac{49}{4}$$, so $$\sqrt[3]{x} = \frac{49}{4} - 12 = \frac{1}{4}$$. Cubing gives $$x = \frac{1}{64}$$, so $$p = 1$$ and $$q = 64$$, which are coprime. Hence $$p + q = 65$$.

A number leaving remainder 1 on division by 2, 3, 4, 5 and 6 is of the form $$60k + 1$$, because the LCM of these divisors is 60. The smallest 4-digit number of this form is $$60 \times 17 + 1 = 1021$$. Its digits are 1, 0, 2 and 1, so it contains only one zero.

Making the value of $$b$$ common in both ratios gives $$a \colon b \colon c = 8 \colon 12 \colon 15$$. Then $$a + c$$ corresponds to $$8 + 15 = 23$$ parts, so one part is $$\frac{736}{23} = 32$$. Hence $$b = 12 \times 32 = 384$$.

In the quadrilateral $$DEFG$$ the angle at $$D$$ is vertically opposite to $$\angle ADC$$ and so equals $$30^\circ$$, giving $$\angle DEF + \angle DGF = 360^\circ - 30^\circ - 120^\circ = 210^\circ$$. Using $$\angle DEF = 2\angle DGF$$ we get $$\angle DEF = 140^\circ$$, so the angle between $$ED$$ and the line $$HEF$$ on the other side is $$40^\circ$$. Applying the angle sum of the quadrilateral $$HBCE$$, $$\angle BHF = 360^\circ - 110^\circ - 80^\circ - 40^\circ = 130^\circ$$.

Combining the left side gives $$\frac{a + b + 2c}{(b+c)(c+a)} = \frac{2}{a+b}$$, so $$(a+b)(a+b+2c) = 2(b+c)(c+a)$$. Expanding both sides gives $$a^2 + 2ab + b^2 + 2ac + 2bc = 2ab + 2bc + 2ac + 2c^2$$, which reduces to $$a^2 + b^2 = 2c^2$$. Hence $$\frac{a^2 + b^2}{c^2} = 2$$.

The whole job equals $$3 \times 43 = 129$$ man-days and also $$4 \times 43 = 172$$ woman-days, so one man does $$\frac{1}{129}$$ and one woman $$\frac{1}{172}$$ of the job each day. Working together, 7 men and 5 women do $$\frac{7}{129} + \frac{5}{172} = \frac{28 + 15}{516} = \frac{43}{516} = \frac{1}{12}$$ of the job per day. Hence the job takes 12 days.

An expression of the form $$u^2(a+b)^2 - v^2(a-b)^2$$ is a difference of two squares, so it factorises as $$[(u - v)a + (u + v)b][(u + v)a + (u - v)b]$$. The four coefficients then add up to $$(u - v) + (u + v) + (u + v) + (u - v) = 4u$$. Here $$u = 7$$, so $$l + m + n + p = 4 \times 7 = 28$$.

Multiplying throughout by 60 clears the fractions and gives $$20(x - 3) - 15(x - 8) = 12(x - 5)$$. Simplifying, $$5x + 60 = 12x - 60$$, so $$7x = 120$$ and $$x = \frac{120}{7} \approx 17.14$$. The integer part of this solution is 17.

The third angle of the triangle is $$180^\circ - 100^\circ - 30^\circ = 50^\circ$$, so the three angles of the triangle add to $$180^\circ$$. At each vertex of the triangle the angles round that point add to $$360^\circ$$, and the polygons contribute their interior angles $$60^\circ$$, $$90^\circ$$ and $$120^\circ$$, each at two vertices. Hence $$x + y + z = 3 \times 360^\circ - 180^\circ - 2(60^\circ + 90^\circ + 120^\circ) = 1080^\circ - 180^\circ - 540^\circ = 360^\circ$$.

The five numbers add up to $$5 \times 105 = 525$$. If the first number is $$f$$, the other four add up to $$525 - f$$, so $$f = \frac{2}{5}(525 - f)$$. This gives $$5f = 1050 - 2f$$, so $$7f = 1050$$ and $$f = 150$$.

If the square has side $$a$$, its area is $$a^2$$ and the new rectangle has sides $$1.3a$$ and $$1.2a$$. Its area is $$1.3a \times 1.2a = 1.56a^2$$. The excess is $$0.56a^2$$, which is 56 % of the original area.

Dividing the equation by $$x$$ gives $$x - \frac{1}{x} = -(2 + \sqrt{3})$$. Squaring both sides, $$x^2 + \frac{1}{x^2} - 2 = (2 + \sqrt{3})^2 = 7 + 4\sqrt{3}$$, so $$x^2 + \frac{1}{x^2} = 9 + 4\sqrt{3}$$. Comparing, $$a = 9$$, $$b = 4$$ and $$c = 3$$, so $$a + b + c = 16$$.

Solving for $$n$$ gives $$n = \frac{44 - m^2}{2m + 1}$$, which must be a positive integer, so $$m$$ can only be 1, 2, 3, 4, 5 or 6. Testing these, only $$m = 2$$ with $$n = 8$$ and $$m = 3$$ with $$n = 5$$ work. The sum of all these values is $$2 + 8 + 3 + 5 = 18$$.

The first bracket equals $$\frac{(a+b)^2 - 4ab}{a+b} = \frac{(a-b)^2}{a+b}$$. Taking $$(a+b)(b-a)$$ as the common denominator, the second bracket equals $$\frac{a(b-a) - b(a+b) + 2ab}{(a+b)(b-a)} = \frac{-(a-b)^2}{(a+b)(b-a)} = \frac{a-b}{a+b}$$. Dividing the first by the second gives $$a - b = 2025 - 2024 = 1$$.

Each term is one less than a perfect cube, since $$0 = 1^3 - 1$$, $$7 = 2^3 - 1$$, $$26 = 3^3 - 1$$, $$63 = 4^3 - 1$$ and $$124 = 5^3 - 1$$. So the $$n$$th term is $$n^3 - 1$$ and the sixth term is $$6^3 - 1 = 216 - 1 = 215$$.

Working from the innermost radical of $$A$$, $$\sqrt{81} = 9$$, then $$\sqrt{112 + 9} = 11$$, then $$\sqrt{53 + 11} = 8$$ and finally $$A = \sqrt{281 + 8} = \sqrt{289} = 17$$. Similarly for $$B$$, $$\sqrt{36} = 6$$, $$\sqrt{75 + 6} = 9$$, $$\sqrt{55 + 9} = 8$$ and $$B = \sqrt{92 + 8} = 10$$. Hence $$A - B = 17 - 10 = 7$$.

The pair averages give $$a + b = 32$$, $$b + c = 52$$ and $$c + a = 50$$, so adding them, $$2(a + b + c) = 134$$ and $$a + b + c = 67$$. Hence $$a = 15$$, $$b = 17$$ and $$c = 35$$. The average of the four numbers is $$b + 4 = 21$$, so $$a + b + c + d = 84$$ and $$d = 84 - 67 = 17$$, making the average of $$d$$ and 67 equal to $$\frac{17 + 67}{2} = 42$$.

The radius $$AB$$ is 10 cm and the two semicircles stand on its two halves, each of length 5 cm, so the two semicircles have exactly the same area. One of them is cut out of the quadrant while the other is added on outside it, so the two changes cancel. The shaded area is therefore the area of the quadrant, $$\frac{1}{4} \times \pi \times 10^2 = 25\pi$$, giving $$k = 25$$.

TO BE FILLED - figure required

Let the tens digit be $$t$$, so the units digit is $$t - 1$$. The condition gives $$(t - 2)(t + 1) = 40$$, that is $$t^2 - t - 42 = 0$$, so $$(t - 7)(t + 6) = 0$$. Only $$t = 7$$ is a digit, so the only such number is 76 and there is exactly 1 such integer.