NMTC KAPREKAR Sub-Junior 2019 Solved Question Paper PDF

Testing digits for $$D$$, only $$D=7$$ gives a product of the required form: $$4921 \times 7 = 34447$$, matching the pattern $$ABBBD$$ with $$A=3$$, $$B=4$$, $$D=7$$. Then $$34447 \times 7 = 241129$$, whose digits sum to $$2+4+1+1+2+9=19$$.

Long division gives $$\frac{5}{28}=0.1\overline{785714}$$, with the first digit 1 followed by a repeating block of 6 digits starting from the second decimal place. Since $$(2019-2) \bmod 6 = 1$$, the 2019th digit matches the 2nd digit of the repeating block "785714", which is 8.

Since $$X$$ has 1000 digits, $$Y \leq 9000$$, so $$Z$$ can be any digit-sum achievable by a number up to 9000, with the highest such digit-sum being 29 (for example when $$Y=2999$$). Taking $$Z=29$$ gives $$W=2+9=11$$, which is the maximum since no value up to 35 (the true maximum of $$Z$$) has a higher digit sum than 29.

Since $$x=10^{-2020}$$ is extremely small, $$10000+x$$ and $$10000-x$$ are both close to 10000, while $$\frac{10000}{x}=10^{2024}$$ is enormous, but $$\frac{1}{x^2}=10^{4040}$$ is far larger still.

The addition reads $$ABC+CBA=DEDD$$, which simplifies to $$101(A+C)+20B=1011D+100E$$. Since the sum is at most 1998, $$D=1$$, and analyzing the equation modulo 20 shows $$A+C$$ must equal 11 for an integer solution, which forces $$E=2$$ (with $$B=5$$ and, for example, $$A=3$$, $$C=8$$).

Placing 0s at just the four cells (2,2), (2,4), (4,2), (4,4) and 1s everywhere else makes every one of the sixteen 2 by 2 sub grids contain exactly one 0, satisfying the condition with the minimum possible number of 0s. This gives a maximum sum of $$25-4=21$$.

Placing $$A=(0,8)$$, $$B=(8,8)$$, $$C=(8,0)$$, $$D=(0,0)$$ gives $$E=(2,8)$$, $$F=(8,6)$$ and $$X=(4,4)$$. Applying the shoelace formula to the quadrilateral $$E, B, F, X$$ gives an area of 16.

Taking the rectangle with width $$w$$ and height $$h$$, the quadrilateral $$AECG$$ has area $$\frac{3wh}{8}$$ by the shoelace formula, while the rectangle has area $$wh$$. So the ratio is $$wh \colon \frac{3wh}{8} = 8 \colon 3$$.

The equation reads $$RAT+MAT+VAT=FLAT$$. Matching units and tens digits forces $$T=0$$ and $$A=5$$, and to maximize the thousands digit, $$R+M+V$$ should be as large as possible using distinct remaining digits while avoiding a units-digit conflict. Taking $$R, M, V = 9, 8, 6$$ gives $$R+M+V=23$$, leading to $$FLAT=2450$$.

The sum of eleven consecutive integers starting at $$n$$ is $$11(n+5)$$, which must be less than 400 and at least $$11 \times 6=66$$ (for $$n=1$$). This gives $$n$$ ranging from 1 to 31, so there are 31 possible sums.

Testing values shows $$xy$$ can exceed 99 (e.g. $$x=y=10.05$$), can be made arbitrarily close to 0, and can equal 75 exactly (e.g. $$x=15, y=5, z=0.1$$), ruling out the first three options. For $$yz=49$$ with $$y \geq z$$, the constraint $$x \geq y$$ combined with $$x+y+z=20.1$$ can be shown to always fail, so $$yz \neq 49$$ always holds.

Computing terms shows the sequence is periodic with period 6, and the sum of any full period is 0. Since $$2019 = 6 \times 336 + 3$$, the total sum equals the sum of the first 3 terms, $$2+4+2=8$$.

The five primes in this range are $$2003, 2011, 2017, 2027, 2029$$. The difference between the largest and smallest is $$2029-2003=26$$.

A pentagon's angles sum to $$540^\circ$$, so 4 right angles would force the fifth to be a degenerate $$180^\circ$$, which is impossible. A parallelogram only has two distinct angle values (equal pairs), so it can never have exactly 3 obtuse angles, and a hexagon with 4 proper reflex angles would already exceed its total angle sum of $$720^\circ$$. The octagon in option (b) has angles summing correctly to $$1080^\circ$$ and is a genuinely constructible equilateral star-like shape.

Since $$2^{10}=1024$$ and $$3^{10}=59049$$, and $$1024 < 2019 < 59049$$, it follows that $$2 < y < 3$$.

There are 9 two-digit terms (11 through 91) and 90 three-digit terms (110-119, 210-219, and so on up to 910-919), together accounting for the first 99 terms. So the 100th term is the smallest four-digit number with second digit 1, which is 1100.

Using the median length formula, $$AD^2=\frac{2AB^2+2AC^2-BC^2}{4}$$ gives $$BC^2=100$$. Since $$AB^2+AC^2=36+64=100=BC^2$$, the triangle is right angled at $$A$$, so its area is $$\frac{1}{2}\times6\times8=24$$.

Combining the two equations eliminates $$b$$ to give $$4a+3c=5$$, and the direction vector of the resulting solution line is $$(-3,-5,4)$$, whose components satisfy $$(-3)^2-(-5)^2+4^2=9-25+16=0$$. This means $$a^2-b^2+c^2$$ is constant along all solutions, and evaluating at one solution (such as $$a=1.25, b=0.75, c=0$$) gives the value 1.

Since $$AD=AE$$, triangle $$ADE$$ is isosceles with base angles $$90-\frac{A}{2}$$, and since $$CE=CF$$, triangle $$CEF$$ is isosceles with base angles $$90-\frac{C}{2}$$. Since $$D$$, $$E$$, $$F$$ lie such that $$A$$, $$E$$, $$C$$ are collinear, $$\angle DEF = 180-(90-\frac{A}{2})-(90-\frac{C}{2})=\frac{A+C}{2}=\frac{90}{2}=45$$.

The smallest 5-digit multiple of 13 is $$13\times770=10010$$ and the largest is $$13\times7692=99996$$. The count of multiples is $$7692-770+1=6923$$.

With angle $$\theta$$ in radians, the sector area is $$\frac{1}{2}r^2\theta$$ and the arc length is $$r\theta$$. Setting these equal and dividing both sides by $$r\theta$$ gives $$\frac{r}{2}=1$$, so $$r=2$$.

Since $$\frac{43}{30}=1+\frac{13}{30}$$, we get $$a=1$$ and the remaining fraction $$\frac{30}{13}=2+\frac{4}{13}$$ gives $$b=2$$. Then $$\frac{13}{4}=3+\frac{1}{4}$$ gives $$c=3$$ and $$d=4$$.

The only candidate ages with two guesses exactly one away are the midpoints of consecutive guesses differing by 2, namely 39, 41, 47 and 58, and among these only 41 and 47 are prime. Since 47 has exactly 5 guesses below it (satisfying "at least half too low") while 41 only has 3, the teacher's age is 47.

Since the LCM is 9, every integer must be a divisor of 9, namely 1, 3, or 9. Writing the counts of each as $$a, b, c$$ with $$a+b+c=8$$ and $$a+3b+9c=22$$ gives the unique solution $$a=4, b=3, c=1$$. So the number of integers less than 4 (the 1s and 3s) is $$4+3=7$$.

Since $$48=2^4 \times 3$$, for $$48n$$ to be a perfect fourth power, $$n$$ must supply a factor of $$3^3$$ times a perfect fourth power, so $$n=27s^4$$ for a positive integer $$s$$. Checking $$n \leq 2019$$ gives $$s=1$$ ($$n=27$$) and $$s=2$$ ($$n=432$$) as the only solutions, so there are 2 such values.