NMTC 2015 Stage II - Gauss (Class 5, 6) - Problems and Solutions

Problem 1

A unit fraction is one of the form $\frac{1}{a}$ where ' $a$ ' is a natural number $(\neq 1)$.Any proper fraction can be written as the sum of two or more unit fractions.
$\left\{\text{Ex.} \frac{1}{2}=\frac{1}{3}+\frac{1}{6}, \frac{5}{6}=\frac{1}{2}+\frac{1}{3}, \frac{1}{24}=\frac{1}{54}+\frac{1}{72}+\frac{1}{108}\right\}$
In the case of $\frac{1}{2}$, the factors of 2 are 1 and 2 only
$$
\frac{1}{2}=\frac{1+2}{2(1+2)}=\frac{1}{2(1+2)}+\frac{2}{2(1+2)}=\frac{1}{6}+\frac{1}{3}
$$

Express $\frac{1}{15}$ as the sum of two different unit fraction in 4 different ways.

Problem 2

(a) The number 11284 and 7655 when divided by a certain 3 digit number leave the same remainder. Find the number and the remainder.
(b) What is the least number to be subtracted from 1936 so that when divided by 9,10 and 15 it will leave the same remainder in each case?

Problem 3

A, M, T, I represent different non-zero digits, It is given that
$$
\begin{aligned}
& A+M+T+I-11 \\
& A+M+I-10 \\
& A+M-1
\end{aligned}
$$

Further in the following addition only one digit is given.

Fill up the stars writing proper reasons.

Problem 4

$A B C$ is a three digt number in which the digit $A$ is greater than the digit $B$ and $C$. If the difference between $A B C$ and $C B A$ is 297 and the difference between $A B C$ and $B A C$ is 450 , find all possible three digit numbers $A B C$ and find their sum also.

Problem 5

Take a triangle. Three straight lines are drawn through the vertices of the triangle as shown. The maximum number of points of intersection is 3 . Draw two lines through each vertex of a triangle to meet the opposite sides. What is the maximum number of points of intersection.

Find again drawing the maximum number of points of intersection when three lines are drawn through each vertex.

Without drawing can you guess the maximum points of intersection for 4 lines?