NMTC 2017 Stage II - KAPREKAR (Class 7, 8) - Problems and Solutions

Problem 1

(a) Find all three digit numbers in which any two adjacent digits differ by 3.
(b) There are 5 cards. Five positive integers (may be different or equal) are written on these cards, one on each card. Abhiram finds the sum of the numbers on every pair of cards. He obtains only three different totals $57,70,83$. Find the largest integer written on a card.

Problem 2

(a) $\mathrm{ABC}$ is a triangle in which $\mathrm{AB}=24, \mathrm{BC}=10$ and $\mathrm{CA}=26 . \mathrm{P}$ is a point inside the triangle. Perpendiculars are drawn to $B C, A B$ and $A C$. Length of these perpendiculars respectively are $x, y$ and $z$. Find the numerical value of $5 x+12 y+13 z$.
(b) If $x^2(y+z)=a^2, y^2(z+x)=b^2, z^2(x+y)=c^2, x y z=abc$ prove that $a^2+b^2+c^2+2 a b c=1$

Problem 3

If $$
\quad \begin{aligned}
X & =\frac{a^2-(2 b-3 c)^2}{(3 c+a)^2-4 b^2}+\frac{4 b^2-(3 c-a)^2}{(a+2 b)^2-9 c^2}+\frac{9 c^2-(a-2 b)^2}{(2 b+3 c)^2-a^2} \\
Y & =\frac{9 y^2-(4 z-2 x)^2}{(2 x+3 y)^2-16 z^2}+\frac{16 z^2-(2 x-3 y)^2}{(3 y+4 z)^2-4 x^2}+\frac{4 x^2-(3 y-4 z)^2}{(4 z+2 x)^2-9 y^2}
\end{aligned}
$$

Find $2017(X+Y)$

Problem 4

The sum of the ages of a man and his wife is six times the sum of the ages of their children. Two years ago the sum of their ages was ten times the sum of the ages of their children. Six years hence the sum of their ages will be three times the sum of the ages of their children. How many children do they have?

Problem 5

(a) $a, b, c$ are three natural numbers such that $a \times b \times c=27846$. If $\frac{a}{6}=b+4=c-4$, find $a+b+c$.
(b) $\mathrm{ABCDEFGH}$ is a regular octagon with side length equal to $a$. Find the area of the trapezium $ABDG$.

Problem 6

(a) If $a, b, c$ are positive real number such that no two of them are equal, show that $a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$ is always positive
(b) In the figure below, $\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}$ are point on the sides of the triangle $\mathrm{ABC}$ such that $\mathrm{CP}=\mathrm{PQ}=\mathrm{QB}=\mathrm{BA}=\mathrm{AR}=\mathrm{RS}=\mathrm{SC}$

NMTC 2017 Stage II - BHASKARA (Class 9, 10) - Problems and Solutions

Problem 1

(a) Find all prime numbers $p$ such that $4 p^2+1$ and $6 p^2+1$ are also primes.
(b) Determine real numbers $x, y, z, u$ such that
$$
\begin{aligned}
& x y z+x y+y z+z x+x+y+z=7 \\
& y z u+y z+z u+u y+y+z+u=9 \\
& z u x+z u+u x+x z+z+u+x=9 \\
& u x y+u x+x y+y u+u+x+y=9
\end{aligned}
$$

Problem 2

If $x, y, z, p, q, r$ are distinct real numbers such that
$$
\begin{aligned}
& \frac{1}{x+p}+\frac{1}{y+p}+\frac{1}{z+p}=\frac{1}{p} \\
& \frac{1}{x+q}+\frac{1}{y+q}+\frac{1}{z+q}=\frac{1}{q} \\
& \frac{1}{x+r}+\frac{1}{y+r}+\frac{1}{z+r}=\frac{1}{r}
\end{aligned}
$$
find the numerical value of $\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)$.

Problem 3

$\mathrm{ADC}$ and $\mathrm{ABC}$ are triangles such that $\mathrm{AD}=\mathrm{DC}$ and $\mathrm{CA}=\mathrm{AB}$. If $\angle \mathrm{CAB}=20^{\circ}$ and $\angle \mathrm{ADC}=100^{\circ}$, without using Trigonometry, prove that $\mathrm{AB}=\mathrm{BC}+\mathrm{CD}$.

Problem 4

(a) a, b, c, d are positive real numbers such that $a b c d=1$. Prove that $$\frac{1+a b}{1+a}+\frac{1+b c}{1+b}+\frac{1+c d}{1+c}+\frac{1+d a}{1+d} \geq 4.$$
(b) In a scalene triangle $\mathrm{ABC}, \angle \mathrm{BAC}=120^{\circ}$. The bisectors of the angles $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ meet the opposite sides in $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ respectively. Prove that the circle on $\mathrm{QR}$ as diameter passes through the point $P$.

Problem 5

(a) Prove that $x^4+3 x^3+6 x^2+9 x+12$ cannot be expressed as a product of two polynomials of degree 2 with integer coefficients.
(b) $2 n+1$ segments are marked on a line. Each of these segments intersects at least $n$ other segments. Prove that one of these segments intersects all other segments.

Problem 6

If $a, b, c, d$ are positive real numbers such that $a^2+b^2=c^2+d^2$ and $a^2+d^2-a d=b^2+c^2+bc$, find the value $$\frac{a b+c d}{a d+b c}$$

NMTC 2017 Stage II - GAUSS (Class 5, 6) - Problems and Solutions

Problem 1

If $$\frac{1}{5 \frac{1}{3}}+\frac{1}{3 \frac{3}{7}}+\frac{1}{4 \frac{4}{7}}+\frac{1}{?}=\frac{77}{96}$$
Find what should be filled in the place marked?

Problem 2

There are $10$ cards numbered $1$ to $10$ . There are three second standard children Ram, Bilal and Cynthia. The teacher selects $3$ cards from the $10$ cards without seeing the numbers. She distributes the cards to the children one to each. After the children noting down the numbers in the cards she collects them back. Again she repeats the same process two more times. So, each child now has $3$ numbers noted down. The teacher asks them to add the numbers and tell her the sums obtained by them. They told her that the sums were $10, 14,15$ But Ram received the same cards three times. Bilal and Cynthia received all cards different. What numbered cards are received by each? Write down the steps you used to get the answer.

Problem 3

In the adjoining figure $ABCD$ is a rectangle. Points $P, Q, R, S$ are marked as in the diagram such that $A P=P Q=Q B . R$ is the midpoint of $C D$. If $A S: S D=3: 1$, find the ratio of the areas of triangle $A S P$, quadrilateral $S P R D$, triangle $P Q R$ and the trapezium $\mathrm{QBCR}$.

Problem 4

Take the numbers $1,2,3,4,5,6,7$ and $8$ . We have to make two groups, $A, B$ each containing four numbers such that
(a) The sum of the numbers in group $A$ is equal to the sum of the numbers in group $B$
(b) Group $A$ has a number such that when it is moved from group A to group B, the sum of the five numbers in group $B$ is equal to twice the sum of the 3 numbers in group $A$.
(c) Group $B$ has a number such that when it moved to group $A$, the sum of the three numbers in Group B is $\frac{5}{7}$ of the sum of the $5$ numbers in Group A.

Find the number in the groups $\mathrm{A}$ and $\mathrm{B}$.

Problem 5

Mahadevan was puzzled by the strange way in which his grand daughter was counting. She began to count on the fingers of her left hand. She started by calling the thumb 1 , the first finger 2 , middle finger 3 , ring finger 4 , little finger 5 , then she reversed direction, calling the ring finger 6 , middle finger 7 , first finger 8 , thumb 9 , then back to the first finger for 10 , middle finger for 11 , and so on. She continued to count back and forth in this peculiar manner until she reached a count of 20 on her ring finger.

Seeing this, Mahadevan told her "If you can find on which finger you will count $2017$ , I will buy you an ice cream". Can you find on which finger she will count $2017$? Explain the steps you used to arrive at the answer.

Problem 6

In the adjoining figure the number in each circle is the sum of the numbers in the two adjacent circles below it.

(a)Find $X$, writing the steps systematically.
(b)What is the least positive number to be added to $X$ so that the result is a perfect square?
(c)What is the least positive number to be subtracted from $X$ so that the result is a perfect square?