$\mathrm{n}, \mathrm{a}$ are natural numbers each greater than 1 . If $a+a+a+a+\ldots+a=2010$, and there are $n$ terms on the left hand side, then the number of ordered pairs $(a, n)$ is
Value of $a$ will be greater than 1. So, first we can find out the factors of 2010.
So,$ 2010= 2\times 3 \times 5\times 67$
When the value of a is 2, value of n is 1005 and
when the value of a is 3, then the value of n is 670 and vice versa.
So, the numbered of ordered pair $(a,n)$ is 14
Sum of the odd numbers from 1 to 2019 both inclusive, is divisible by
This can be solved using Arithmetic progression
Where $S= 1+3+5+7+\dots+ 2019 $
Then use the formula of sum of Arithmetic Progression
As for example, $Sum$= $101\times 101 \times 100$
Divisible by both 100 and 101
$X$ is a 5 digit number. Let $Y$ be the sum of the digits of $X$. Let $Z$ be the sum of the digits of $Y$. Then the maximum possible value that $Z$ can have is
If $x$= 99999
$y$= 9+9+9+9+9=45
$z$= 4+5=9 That is not maximum
$Y$ must have a digit 9, to get the maximum value of z
By Trial method $x$= 99993
So the answer is 12
Look at the set of numbers ${2,3,5,7,8,10,12}$. Four numbers are selected from this and made into two pairs. The pairs are added and the resulting two numbers are multiplied. The smallest such product is
For smallest product , we need to select smallest four numbers
From the set $\{2,3,5,7,8,10,12\}$
The smallest numbers are $\{2,3,5,7\}$
Pairs may be $\{2,3\}$ & $\{5,7\}$
$\{2,5\}$ & $\{3,7\}$ ..............etc
So smallest such product is 60
A calendar for 2019 is made using 4 sheets, each sheet having 3 months. The total number of days shown in each of the four sheets $\left(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {red }}, 4^{\text {th }}\right)$ respectively is
Four sheets are made like this
1st sheet $\{ Jan, Feb, March\}$
Now add the numbers of days of four sheets
For 1st sheet $\{Jan, Feb, March\}$
$\{31+28+31\}$= $\{90\}$
Now adding total no of days shown in each of the four sheets
we get the answer $\{90,91,92,92\}$
Triples of odd numbers $(a, b, c)$ with $a<b<c$, with $a, b, c$ from 1 to 10 are generated such that $a+b+c$ is prime number. The number of such triple is
Find the odd numbers from 1 to 10
Then add any three numbers from the set of odd numbers and check the result is
prime or not.
Now by trial method, we get 6 number of such triple
Given below is the triangular form of AMTI.
$\mathrm{A}$
A M A
A M TM A
A M T I T M A
The number of ways you can spell AMTI, top to bottom, right to left or left to right or a combination of these is
Hints and solutions are coming up soon.
We can read AMTI by following ways
$\{1,3,7,13\}$ $\{5,6,12,13\}$............etc
Thus total no of ways we get is 15
A string of beads has a recurring pattern as follows: 5 blue, 4 black, 4 white, 5 blue, 4 black, 4 white ……….. and so on. The colour of the $321^{\text {st }}$ bead is
Pattern is 5 Blue, 4 Black, 4 White
$ 5+4+4 = 13 $
Now, $321= 13\times 24+ 9$
As 9= 5+4
So $321^{\text{st}}$ bead is Black
In a $5 \times 5$ grid having 25 cells, Janani has to enter 0 or 1 in each cell such that each sub square grid of size $2 \times 2$ has exactly three equal numbers. What is the maximum possible sum of the numbers in all the 25 cells put together?
For maximum possible sum, we need to enter more 1 less 0
In $2\times 2$, square has exactly three 1 and one 0
Maximum possible sum $1+1+1+1+1+\dots+1=21$
Numbers of 5 -digit multiples of 13 is
Smallest 5-digit multiple of 13 is =$ 10010$
= $770\times 13$
Largest 5 digit multiple of 13 is = $99996$
=$7692\times 13$
Now we can write by following way,
$(770\times 13), (771\times 13)\dots (7692\times 13)$
so the answer is 6923
In a room, $50 \%$ of the people are wearing gloves, and $80 \%$ of the people are wearing hats. The minimum percentage of people in the room wearing both a hat and a glove is
Hints and solutions are coming up soon.
$x+y= 50%\dots$ (i)
$y+z=80%\dots$(ii)
$x+y+z=100%\dots$(iii)
From (i)+(ii)-(iii), we get the answer is 30%
Given a sheet of 16 stamps as shown, the number of ways of choosing three connected stamps (two adjacent stamps must have an edge in common) is
For each $2\times 2 $ block , we can select 3 stamps by 4 ways
Now, $1\times 3$ block, we can select 3 stamps by 1 way
Required no of ways= 28+12+2=42
A $4 \times 4$ anti-magic square is an arrangement of the numbers 1 to 16 in a square so that the totals of each of the four rows, four columns and the two diagonals are ten consecutive numbers in some order. The diagram shows an incomplete anti magic square. When it is completed, the number in the position of ${ }^{*}$ is
Sum would be 30 to 39 ( 10 consecutive no)
Remaining sums are $\{32, 33, 34, 35, 36, 37, 38\}$
Remaining numbers to be filled are 1, 2, 8, 15, 16
So the number is 16
In a stack of coins, each row has exactly one coin less than the row below. If we have nine coins, two such towers are possible. Of these, the tower on the left is the tallest. If you have 2015 coins, the height of the tallest towers is
Take a possibility such that the topmost row will start with 2 coins.
Take a variable which represents that height of the tower
Use the sum of A.P. series to find the height of the tower.
So the answer is 62.
In a single move a King $\mathrm{K}$ is allowed to move to any of the squares touching the square it is on, including diagonals, as indicated in the figure. The number of different paths using exactly seven moves to go from $A$ to $B$ is
First find the minimum no of moves to go from A to B
King can move to at most 3 square from the previous square
Find all the possible square the king can move if the king has to complete it in exactly
7 moves.
The answer is 127
$\mathrm{n}, \mathrm{a}$ are natural numbers each greater than 1 . If $a+a+a+a+\ldots+a=2010$, and there are $n$ terms on the left hand side, then the number of ordered pairs $(a, n)$ is
Value of $a$ will be greater than 1. So, first we can find out the factors of 2010.
So,$ 2010= 2\times 3 \times 5\times 67$
When the value of a is 2, value of n is 1005 and
when the value of a is 3, then the value of n is 670 and vice versa.
So, the numbered of ordered pair $(a,n)$ is 14
Sum of the odd numbers from 1 to 2019 both inclusive, is divisible by
This can be solved using Arithmetic progression
Where $S= 1+3+5+7+\dots+ 2019 $
Then use the formula of sum of Arithmetic Progression
As for example, $Sum$= $101\times 101 \times 100$
Divisible by both 100 and 101
$X$ is a 5 digit number. Let $Y$ be the sum of the digits of $X$. Let $Z$ be the sum of the digits of $Y$. Then the maximum possible value that $Z$ can have is
If $x$= 99999
$y$= 9+9+9+9+9=45
$z$= 4+5=9 That is not maximum
$Y$ must have a digit 9, to get the maximum value of z
By Trial method $x$= 99993
So the answer is 12
Look at the set of numbers ${2,3,5,7,8,10,12}$. Four numbers are selected from this and made into two pairs. The pairs are added and the resulting two numbers are multiplied. The smallest such product is
For smallest product , we need to select smallest four numbers
From the set $\{2,3,5,7,8,10,12\}$
The smallest numbers are $\{2,3,5,7\}$
Pairs may be $\{2,3\}$ & $\{5,7\}$
$\{2,5\}$ & $\{3,7\}$ ..............etc
So smallest such product is 60
A calendar for 2019 is made using 4 sheets, each sheet having 3 months. The total number of days shown in each of the four sheets $\left(1^{\text {st }}, 2^{\text {nd }}, 3^{\text {red }}, 4^{\text {th }}\right)$ respectively is
Four sheets are made like this
1st sheet $\{ Jan, Feb, March\}$
Now add the numbers of days of four sheets
For 1st sheet $\{Jan, Feb, March\}$
$\{31+28+31\}$= $\{90\}$
Now adding total no of days shown in each of the four sheets
we get the answer $\{90,91,92,92\}$
Triples of odd numbers $(a, b, c)$ with $a<b<c$, with $a, b, c$ from 1 to 10 are generated such that $a+b+c$ is prime number. The number of such triple is
Find the odd numbers from 1 to 10
Then add any three numbers from the set of odd numbers and check the result is
prime or not.
Now by trial method, we get 6 number of such triple
Given below is the triangular form of AMTI.
$\mathrm{A}$
A M A
A M TM A
A M T I T M A
The number of ways you can spell AMTI, top to bottom, right to left or left to right or a combination of these is
Hints and solutions are coming up soon.
We can read AMTI by following ways
$\{1,3,7,13\}$ $\{5,6,12,13\}$............etc
Thus total no of ways we get is 15
A string of beads has a recurring pattern as follows: 5 blue, 4 black, 4 white, 5 blue, 4 black, 4 white ……….. and so on. The colour of the $321^{\text {st }}$ bead is
Pattern is 5 Blue, 4 Black, 4 White
$ 5+4+4 = 13 $
Now, $321= 13\times 24+ 9$
As 9= 5+4
So $321^{\text{st}}$ bead is Black
In a $5 \times 5$ grid having 25 cells, Janani has to enter 0 or 1 in each cell such that each sub square grid of size $2 \times 2$ has exactly three equal numbers. What is the maximum possible sum of the numbers in all the 25 cells put together?
For maximum possible sum, we need to enter more 1 less 0
In $2\times 2$, square has exactly three 1 and one 0
Maximum possible sum $1+1+1+1+1+\dots+1=21$
Numbers of 5 -digit multiples of 13 is
Smallest 5-digit multiple of 13 is =$ 10010$
= $770\times 13$
Largest 5 digit multiple of 13 is = $99996$
=$7692\times 13$
Now we can write by following way,
$(770\times 13), (771\times 13)\dots (7692\times 13)$
so the answer is 6923
In a room, $50 \%$ of the people are wearing gloves, and $80 \%$ of the people are wearing hats. The minimum percentage of people in the room wearing both a hat and a glove is
Hints and solutions are coming up soon.
$x+y= 50%\dots$ (i)
$y+z=80%\dots$(ii)
$x+y+z=100%\dots$(iii)
From (i)+(ii)-(iii), we get the answer is 30%
Given a sheet of 16 stamps as shown, the number of ways of choosing three connected stamps (two adjacent stamps must have an edge in common) is
For each $2\times 2 $ block , we can select 3 stamps by 4 ways
Now, $1\times 3$ block, we can select 3 stamps by 1 way
Required no of ways= 28+12+2=42
A $4 \times 4$ anti-magic square is an arrangement of the numbers 1 to 16 in a square so that the totals of each of the four rows, four columns and the two diagonals are ten consecutive numbers in some order. The diagram shows an incomplete anti magic square. When it is completed, the number in the position of ${ }^{*}$ is
Sum would be 30 to 39 ( 10 consecutive no)
Remaining sums are $\{32, 33, 34, 35, 36, 37, 38\}$
Remaining numbers to be filled are 1, 2, 8, 15, 16
So the number is 16
In a stack of coins, each row has exactly one coin less than the row below. If we have nine coins, two such towers are possible. Of these, the tower on the left is the tallest. If you have 2015 coins, the height of the tallest towers is
Take a possibility such that the topmost row will start with 2 coins.
Take a variable which represents that height of the tower
Use the sum of A.P. series to find the height of the tower.
So the answer is 62.
In a single move a King $\mathrm{K}$ is allowed to move to any of the squares touching the square it is on, including diagonals, as indicated in the figure. The number of different paths using exactly seven moves to go from $A$ to $B$ is
First find the minimum no of moves to go from A to B
King can move to at most 3 square from the previous square
Find all the possible square the king can move if the king has to complete it in exactly
7 moves.
The answer is 127
