Test of Mathematics Solution Subjective 181 - Diagonal Moves

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 181 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Suppose that one moves along the points (m, n ) in the plane where m and n are integers in such a way that each move is a diagonal step, that is, consists of one unit to the right or left followed by one unit either up or down.

(a) Which point (p, q) can be reached from the origin.

(b) What is the minimum number of moves needed to reach such a point (p, q)?


Discussion

For each horizontal movement (right or left) there is one vertical movement (up or down). Suppose there are R right moves, L left moves. Also there be U upward move and D downward move. Since number of horizontal movement equals number of vertical movements hence we have

R+L = U+D = k (say)

The final coordinate reached is (R - L , U - D) (as each right move counts as +1 and left move counts as -1, and similarly for vertical movements),

L = k -R

D = k -U

Hence the final coordinate is (2R - k, 2U - k)

Therefore both the coordinates are of same parity (that is either both are even or both are odd).

So we have shown that the final point that can be reached has both x and y coordinate of the same parity. Now we show the converse. That is all the points whose x and y coordinates are of same parity can be reached.

Say P be a point whose both coordinates are even. Without loss of generality we may say P = (2m, 2n) where $ 2m \ge 2n $ and both coordinates are positive.

1. 2(m-n) moves of the form Right-Up followed by Left-down is performed to reach the point (2(m-n) , 0)

2. Next 2n moves of the form Right-Up is performed to reach the point (2m, 2n)

So in total 2(m-n) + 2n = 2m moves are performed to reach the point (2m, 2n)

Similarly to reach a point whose both coordinates are odd, say (2j+1, 2l + 1),  we first reach the point (2j, 2l) and then move one more unit diagonally upward.

This same algorithm will work for all the quadrants.

Also if (p, q) is the final coordinate with p > q, we have taken p steps to reach the point. It is not possible to reach the point in less than p steps, because we must cover at least p boxes in at least one direction and in one step exactly one box can be covered.

Therefore we have found the answer:

(1) All the points (p, q) can be reached where $ p \equiv q \text{mod} 2 $

(2) If $ p \ge q$ , minimum number of steps required is p. If $ q \ge p $, the minimum number of steps required is q.

Test of Mathematics Solution Subjective 83 - Two numbers adding up to 1

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 83 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


Problem

If a and b are positive real numbers such that a + b = 1, prove that $ \displaystyle \left( a + \frac{1}{a} \right)^2 $ + \( \left( b + \frac{1}{b} \right)^2 \ge \frac{25}{2} \)


Solution:

We replace 1 by a+b and expand the squares.

$ \displaystyle \left( a + \frac{a+b}{a} \right)^2 + \left(b + \frac{a+b}{b} \right)^2 $
$ \displaystyle \left( a + \frac{b}{a} + 1 \right)^2 + \left(b + \frac{a}{b} + 1 \right)^2 $
$ \displaystyle \left( a^2 + \frac{b^2}{a^2} + 1 + 2a + 2b + 2\frac{b}{a} \right) + \left(b^2 + \frac{b^2}{a^2} + 1 + 2b + 2a + 2 \frac{a}{b} \right) $

Now we use A.M. - G.M. inequality to have

$ \displaystyle { \frac {\frac{b^2}{a^2} + \frac{a^2}{b^2}}{2}\ge \sqrt {\frac{b^2}{a^2} \times \frac{a^2}{b^2}} = 1 } $
$ \displaystyle { \frac {\frac{b}{a} + \frac{a}{b}}{2}\ge \sqrt {\frac{b}{a} \times \frac{a}{b}} = 1 } $

Also we apply the square mean inequality to have

$ \displaystyle {\frac{a^2 + b^2}{2} \ge \left(\frac{a+b}{2} \right)^2} $ Since a+b = 1, we have $ a^2 + b^2 \ge \frac{1}{2} $

Combining all of them we have $ \displaystyle {\left( a + \frac{1}{a} \right)^2 + \left(b + \frac{1}{b} \right)^2 \ge 12 \frac{1}{2} = \frac{25}{2}} $


Discussion

Some theoretical background.

Clocky Rotato Arithmetic

Do you know that CLOCKS add numbers in a different way than we do? Do you know that ROTATIONS can also behave as numbers and they have their own arithmetic? Well, this post is about how clock adds numbers and rotations behave like numbers. Let's learn about clock rotation today

Consider the clock on earth.

clock

So, there are 12 numbers {1,2, ..., 12 } are written on the clock. But let's see how clocks add them.

What is 3+ 10 ?

Well, to the clock it is nothing else than 1. Why?

Say, it is 3 am and the clock shows 3 on the clock. Now you add 10 hours to 3 am. You get a 13th hour of the day. But to the clock, it is 1 pm.

So, 3 + 10 = 1.

If you take any other addition, say 9 + 21 = 6 to the clock ( 9 am + 21 hours = 6 pm ).

Now, you can write any other Clocky addition. But you will essentially see that the main idea is :

The clock counts 12 = 0.

Isn't it easy? 0 comes as an integer just before 1, but on the clock, it is 12 written. So 12 must be equal to 0. Yes, it is that easy.

Cayley's Table

This is a handsome and sober way to write the arithmetic of a set. It is useful if the set is finite like the numbers of the CLOCK Arithmetic.

Let me show you by an example.

Consider the planet Cheenta. A day on Cheenta consists of 6 earth hours.

So, how will the clock on Cheenta look like?

Cheenta Planet clock

Let's us construct the Cayley Table for Cheenta's Clocky Arithmetic. Check it really works as you wish. Here for Cheenta Clock, 3 = 0.

Clock rotation explanation

Exercise: Draw the Cayley Table for the Earth (24 hours a day) and Jupiter (10 hours a day).

Nice, let's move on to the Rotato part. I mean the arithmetic of Rotation part.

Let's go through the following image.

clock rotation

Well, let's measure the symmetry of the figure. But how?

Well, which is more symmetric : The Triskelion or the Square (Imagine).

clock rotation

Well, Square seems more right? But what is the thing that is catching our eyes?

It is the set of all the symmetric positions, that capture the overall symmetry of a figure.

For the Triskelion, observe that there are three symmetric operations that are possible but that doesn't alter the picture:

For the Square, the symmetries are:

For, a square there are symmetries, hence the eyes feel that too.

So, what about the arithmetic of these? Let's consider the Triskelion.

Just like 1 interact (+) 3 to give 4.

We say \(r_1\) interacts with \(r_2\) if \(r_1\) acts on the figure after \(r_2\) i.e ( 240 + 120 = 360 degrees rotation = \(r_3\) ).

Hence, this is the arithmetic of the rotations. To give a sober look to this arithmetic, we draw a Cayley Table for this arithmetic.

Treskelion

Well, check it out.

Exercise: Can you see any similarity of this table with that of anything before?

Challenge Problem: Can you draw the Cayley Table for the Square?

You may explore this link:- https://www.cheenta.com/tag/level-2/

And this video:- https://www.youtube.com/watch?v=UaGsKzR_KVw

Don't stop investigating.

All the best.

Hope, you enjoyed. 🙂

Passion for Mathematics.

Prime numbers in A.P. | TOMATO Objective 152

If three prime numbers, all greater than $3$, are in A.P. , then their common difference

(A) must be divisible by $2$ but not necessarily by $3$;

(B) must be divisible by $3$ but not necessarily by $2$;

(C) must be divisible by both $2$ and $3$;

(D) need not be divisible by any of $2$ and $3$;

Discussion:

Say $p, p+d$ and $p+2d$ are the three primes in A.P. with d as the common difference.

Since $p>3$, hence $p$ is odd.

If $d$ is odd, then $p+d$ is even. But then $p+d$ cannot be a prime any more. Hence contradiction. Thus $d$ cannot be odd. Hence $2$ divides $d$.

Again $p> 3$ implies $p$ is not divisible by $3$. Hence p is either $1$ or $2$ modulo $3$. Now we wish to check if d is divisible by $3$ or not. Suppose it is not then d is also either $1$ or $2$ modulo $3$.

Case 1: $p$ is $1$ mod $3$ and $d$ is $1$ mod $3$, then $p+2d$ is $0$ mod $3$ hence contradiction (as p+2d is a prime greater than 3).

Case 2: $p$ is $1$ mod $3$ and $d$ is $2$ mod $3$, then $p+d$ is $0$ mod $3$ hence contradiction (as p+d is a prime greater than 3)

Case 1: $p$ is $2$ mod $3$ and $d$ is $1$ mod $3$, then $p+d$ is $0$ mod $3$ hence contradiction (as p+d is a prime greater than 3).

Case 2: $p$ is $2$ mod $3$ and $d$ is $2$ mod $3$, then $p+2d$ is $0$ mod $3$ hence contradiction (as p+2d is a prime greater than 3)

Hence $d$ must be $0$ modulo $3$.

Therefore common difference must be divisible by both $2$ and $3$.

Answer: (C)

Some Useful Links:

Solving a few Diophantine Equations – Video

ISI 2015 BStat – BMath Objective Problems

TOMATO Objective 153 | ISI Entrance | N! -1

Let N be a positive integer not equal to 1. Then note that none of the numbers 2, 3, ... , N is a divisor of (N! -1). From this we can conclude that:

(A) (N! - 1) is a prime number;

(B) at least one of the numbers N+1 , N+2 , ...., N! - 2 is a divisor of (N! -1);

(C) the smallest number between N and N! which is a divisor of (N!-1), is a prime number;

(D) none of the foregoing statements is necessarily correct;

Discussion:

N! - 1 could be a prime (example N = 4 gives N! - 1 = 23). It might not be a prime as well (example N = 5 gives N! - 1 = 119 which is divisible by 7).

Hence none of option A or B is necessarily true.

However option (C) is true as N! - 1 is not a prime, it's first prime divisor occurs between N and N! (since none of the primes from 1 to N divides it. Even if it is a prime, the smallest number between N and N! that divides it is itself which is a prime.

Answer: C

Number of zeroes after factorial |TOMATO Objective 154

The number $1000! = 1.2.3...1000$ ends exactly with

(A) $249$ zeroes;

(B) $250$ zeroes;

(C) $240$ zeroes;

(D) $200$ zeroes;
Discussion:

To find the number of zeroes at the end of n! we just need to figure out the number of 5's occurring in prime factorization of it.  Why? Because there are much more 2's present than there are 5's (every second number is even hence contributing at least one 2 to the prime factorization). Every 0 at the end of n! is basically one 10 made of one 2 and one 5.

Every fifth number contributes at least one 5.  This allows for $\left \lfloor \frac {1000}{5} \right \rfloor $ = 200 fives.

Again every 25th number contributes an extra 5. This allows for $\left \lfloor \frac {1000}{25} \right \rfloor $ = 40 extra fives.

Every 125th number contributes another extra 5. This allows for $\left \lfloor \frac {1000}{125} \right \rfloor $ = 8 more extra fives.

Finally every 625th number contribute one more extra 5. This allows for $\left \lfloor \frac {1000}{625} \right \rfloor $ = 1 more extra fives.

Hence there are total 249 fives in the prime factorization of 1000!. Hence it ends with 249 zeroes.

Answer: (A)