$\mathrm{n}, \mathrm{a}$ 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 ordered pairs will be, $(2,1005), (3,670), (5, 402), (6, 335), (10, 201), (15, 134), (30, 67), (67,30), (134, 15), (201, 10), (6, 335), (402, 5), (670, 3), (1005, 2)$
So the number of ordered pairs $(a, n)$ is $14$.
The number of times the digit occurs in the result of $1+11+111+\ldots .+111$
111 (100digits) is $\ldots \ldots .$
sum is
$(123456790) (123456790)\ldots (123456790) (1234567890)$
there are $11$ brackets
so $1$ comes $11$ times
Let us call a sum of integers a cool sum if the first and last terms are 1 and each term differs from its neighbours by at most. For example, $1+2+2+3+3+2+1$ and $1+2+3+4+3+2+1$ are cool sums. The minimum number of terms required to write 2019 as a cool sum is ……
Sequence is
$$
\begin{aligned}
1+2+3+\ldots . .+19+20+21 &+22+23+\ldots \ldots +39+39+40+41+42+43+44+44+43+42+\ldots \ldots+3+2+1
\end{aligned}
$$
So minimum number of terms is $89$.
For each positive integer $n$ let $f(n)=n^{4}-3 n^{2}+9$. Then the sum of all $f(n)$ which are prime is
$f(n)=n^{4}-3 n^{2}+9$
$=n^{4}+6 n^{2}-9 n^{2}+9$
$=\left(n^{2}+3\right)^{2}-(3 n)^{2}$
$=\left(n^{2}+3 n+3\right)\left(n^{2}-3 n+3\right)$
$f(n)$ is prime $\Rightarrow n^{2}-3 n+3=1 \Rightarrow n=1,2$
$f(1)=7, f(2)=13$
sum of values $=7+13=20$
The product of four positive integers $a, b, c$ and $d$ is 9 ! The number $a, b, c, d$ satisfy $a b+a+b=$ 1224, $b c+b+c=549$ and $c d+c+d=351$. The $a+b+c+d=\ldots \ldots$
$$
\text { abcd }=\angle 9=9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
$$
given $a b+a+b=1224 \quad \Rightarrow(1+a)(1+b)=49 \times 25=1225$
Therefore $b c+b+c=549 \quad \Rightarrow(1+b)(1+c)=25 \times 22=550$
and $c d+c+d=351 \Rightarrow(1+c)(1+d)=352=22 \times 16$
So $1+a=49,1+b=25,1+c=22,1+d=16$
$$
a=48, b=24, c=21, d=15
$$
The number of 6 digit numbers of the form "ABCABC", which are divisible by 13 , where $A, B$ and $C$ are distinct digits, $A$ and $C$ being even digits is
$1001 \times \mathrm{ABC}=\mathrm{ABCABC}$ where $1001=13 \times 7 \times 11$ Now $\mathrm{A}$ and $\mathrm{C}$ are even digits and $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are different digits .
Case-I: When $\mathrm{C}$ is zero
Case-II : When $\mathrm{C}$ is not zero
Total number of $6$ digits
Number possible $= 32 + 96 = 128$
In the sequence $1,4,8,10,16,21,25,30$ and 43 , the number of blocks of consecutive terms whose sums are divisible by 11 is
$4+8+10=22$
$8+10+16+21=55$
$8+10+16+21+25+30=110$
$25+30=55$
Exactly four
In the subtraction below, what is the sum of the digits in the result ? $111 \ldots \ldots \ldots . .111$ (100 digits) $-222 \ldots . . .222$ (50 digits)
$1111 \ldots \ldots \ldots \ldots \ldots \ldots .111 \ldots \ldots \ldots \ldots 111$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 222 \ldots \ldots \ldots \ldots .222$
---------------------------------------------
$1111 \ldots \ldots \ldots \ldots \ldots \ldots .108 \ldots \ldots \ldots \ldots .889$
49 times 1,49 times 8 and 1 times 0 and 9
Sum $=49 \times 1+8 \times 49+9$
Therefore
$\Rightarrow \quad 49+392+9=450 .$
The least odd prime factor of $2019^{8}+1$ is
Let $P$ be an odd prime which divides $2019^{8}+1$
So $2019^{8} \equiv-1(\bmod P)$
$\Rightarrow \quad 2019^{16} \equiv 1(\bmod P)$
Now by Euler's theorem
$2019^{P-1} \equiv 1(\bmod \mathrm{P})$
So $P-1$ should be divisible by $16$
Where $P$ is a prime
First two prime numbers which gives remainder $1$ when divided by $16$ is $17$ and $97$
Case-1 $\quad \mathrm{P}=17$
$2019^{8}+1 \equiv 13^{8}+1 \equiv 4^{8}+1 \equiv 16^{4}+1 \equiv 2 $(mode $17$)
While
$2019^{8}+1=79^{8}+1=18^{8}+1=324^{4}+1=33^{4}+1=1089^{2}+1=22^{2}+1=485=0$ (mod $97$)
So the answer is $97$ .
Let $a, b, c$ be positive integers each less than 50 , such that $a^{2}-b^{2}=100 c$. The number of such triples $(a, b, c)$ is
$a^{2}-b^{2}=100 c$
As $a^{2}-b^{2}$ is a multiple of $100 .$
So it means the last $2$ digit of $a^{2}$ and $b^{2}$ is same.
So $(a, b)$ can be $(49,1)(48,2)(47,3) \ldots \ldots \ldots \ldots .(26,24)$ .
So there are $24$ such pairs .
One more pair for $(a, b)$ is $(25,15)$ .
So total $25$ pairs are possible.
If $4921 \times D=A B B B D$, then the sum of the digits of $A B B B D \times D$ is
4-digit no. (4921) is multiplied by a single digit no. (D) \& result is five digit no., so definitely $D>2$
So by hit \& trial we put the values of D from 3 to 9 .
at $\mathrm{D}=7$
$4921 \times 7=34447$
$(\mathrm{ABBBD})$
So $\mathrm{A}=3, \quad \mathrm{~B}=4, \mathrm{D}=7$
Now ABBBD $(34447) \times 7=241129$
Sum of digits $=2+4+1+1+2+9=19$
What is the $2019^{\text {th }}$ digit to the right of the decimal point, in the decimal representation of $\frac{5}{28}$ ?
$$
\frac{5}{28}=\cdot 17 \overline{857142}
$$
$\Rightarrow \quad 2019=2+336 \times 6+1$ [2 for $17 \& 336$ pairs of 6 repeating number]
$2019^{\text {th }}$ digit from right side to decimal is first digit in repetition
So correct answer is $8$.
If $\mathrm{X}$ is a 1000 digit number, $Y$ is the sum of its digits, $Z$ the sum of the digits of $Y$ and $W$ the sum of the digits of $Z$, then the maximum possible value of $W$ is
$\mathrm{X} \rightarrow 1000$ digit no. If all digit are ' 9 ' so that maximum sum of digit of ' $\mathrm{X}$ ' is 9000 So maximum value of $Y$ is 9000
But for maximum sum of digit of $Y$ is 35 for number (8999) So $Z$ is maximum 35 .
Now for maximum sum of digit of $Z$ is 11 for number $29 .$ So $W=11$. Practical example: if $\quad X=\underbrace{99 \ldots 9}_{333 \text { times }} \quad \underbrace{000 \ldots 0}_{666 \text { times }} 2$ Sum of digit of $X=Y=2999$ Sum of digit of $Y=Z=29$ Sum of digit of $Z=W=11$
Let $x$ be the number $0.000$. 001 which has 2019 zeroes after the decimal point. Then which of the following numbers is the greatest?
(A) $10000+x$, (B) $10000 \cdot x$, (C) $\frac{10000}{x}$, (D) $\frac{1}{x^{2}}$
$$
x=\cdot 000 \ldots 01=10^{-2020}
$$
From option $(A)=10000+x=1000+10^{-2020}=10000 \cdot \underbrace{000 \ldots .01}_{2019 \text { times }}$
From option $(\mathrm{B})=10000 \times \mathrm{x}=10^{4} \times 10^{-2020}=10^{-2016}$
From option $(C)=\frac{10000}{x}=\frac{10^{4}}{10^{-2020}}=10^{2024}$
From option $(D)=\frac{1}{x^{2}}=\frac{1}{\left(10^{-2020}\right)^{2}}=10^{4040}$
So from options $\frac{1}{x^{2}}$ is greatest.
$A B C$
then the number of possible values for $A, B, C, D, E$ satisfying this equation where
If $\frac{C \mathrm{CBA}}{\mathrm{DEDD}}$ then the number of $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ and $\mathrm{E}$ are distinct digits is
$\quad \quad ABC$
$+\quad CBA$
_________
$\quad DEDD$
$D \rightarrow$ must be $1$
Means $C+A=1$ or 11
$So \quad$ Sum of $B+B$ is even $\& D$ is $1 .$
So possible values of B is 0 or 5 .
But if we take $B$ as ' 0 ' so there is no carry forward \& Sum of $A \& C$, did not get different digit from D.
So B must be 5 .
Sum is convert into
$A 5 C$
$C 5 A$
_____
$1211$
Now possible pairs of $(A, C)$ are $(3,8)(4,7)(7,4)(8,3)$
So total 4 possible solutions are there.
$\mathrm{R} A \mathrm{~T}$
If $+\mathrm{MAC}$ and each alphabet represents a different digit, what is the maximum possible value $+\frac{\mathrm{VAT}}{\mathrm{FLAT}}$
of FLAT?
T should be 0 or 5
But if we take $T$ as ' 5 ' sum is $15 \& 1$ carry is forward and sum of 3 'A' and '1' never give unit digit 'A' So T must be '0'
Now again possible values of $A$ are $' 0 ' \& 5$ but alphabet represents different digits so $\mathrm{A}$ is 5 . For maximum value of FLAT, we take maximum value of $R, M \& V$ as $9,8 \& 7$, but sum is $25 \& 5$ is repeat.
So by taking $R, M \& V$ as $9,8 \& 6$. We get maximum value of FLAT.
$$
$\ \ 9 5 0 $
$\ \ 8 5 0 $
$\ \ 6 5 0 $
______
$ 2 4 5 0 $
$FLAT \Rightarrow 2450$
There are exactly 5 prime numbers between 2000 and 2030 . Note: $2021=43 \times 47$ is not a prime number. The difference between the largest and the smallest among these is
Prime no's between $2000 \& 2030$ are $2003,2011,2017,2027,2029$
Difference between $2029 \& 2003$ is
$$
2029-2003=26
$$
A teacher asks 10 of her students to guess her age. They guessed it as $34,38,40,42,46,48,51$, 54,57 and 59 . Teacher said "At least half of you guessed it too low and two of you are off by one. Also my age is a prime number". The teacher's age is
Age of teacher is greater than $46$ .
Again according to questions two of them are off by one.
So there are two possibilities $47 \&\ 58$
$47$ from $(46 \&\ 48)$
$58$ from $(57 \&\ 59)$
But $58$ is not a prime so age of teacher in $47$ .
The sum of 8 positive integers is 22 and their LCM is 9 . The number of integers among these that are less than 4 is
L.C.M. is 9 , so all numbers are from $1,3 \&\ 9$. If we take 2 times 9 sum as 22 is not possible so, 9 will come only one time $\&$ remaining 7 numbers are from $1 \&\ 3$.
So 4 times $1 \&\ 3$ times 3 will come.
$$
1+1+1+1+3+3+3+9=22
$$
So numbers less than 4 is 7
The number of natural numbers $n \leq 2019$ such that $\sqrt[3]{48 n}$ is an integer is
So $\mathrm{n}$ should be multiple of $2^{2} \times 3^{2}=36$ for integer value.
$\mathrm{n} \leq 2019$ $\sqrt[3]{48 \mathrm{n}}=$ integer $\sqrt[3]{2^{4} \times 3 \times n}=$ integer $2 \times \sqrt[3]{2 \times 3 n}=$ integer.
So $\mathrm{n}$ should be multiple of $2^{2} \times 3^{2}=36$ for intege So, possible 'n' are $=2^{2} \times 3^{2}=36<2019$
$=2^{2} \times 3^{2} \times 2^{3}=288<2019$
$=2^{2} \times 3^{2} \times 3^{3}=972<2019$
$=2^{2} \times 3^{2} \times 4^{3}=2304>2019$ (reject)
So only three values of $n$ are possible $36,288,972$
The number of perfect cubes that lie between $2^{9}+1$ and $2^{18}+1$ is
$\left(2^{3}\right)^{3}+1$ and $\left(2^{6}\right)^{3}+1$ $8^{3}+1$ and $64^{3}+1$
Numbers lies between $9,10,11, \ldots \ldots \ldots .64$ Total perfect cubes number $64-8=56$.
$\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 ordered pairs will be, $(2,1005), (3,670), (5, 402), (6, 335), (10, 201), (15, 134), (30, 67), (67,30), (134, 15), (201, 10), (6, 335), (402, 5), (670, 3), (1005, 2)$
So the number of ordered pairs $(a, n)$ is $14$.
The number of times the digit occurs in the result of $1+11+111+\ldots .+111$
111 (100digits) is $\ldots \ldots .$
sum is
$(123456790) (123456790)\ldots (123456790) (1234567890)$
there are $11$ brackets
so $1$ comes $11$ times
Let us call a sum of integers a cool sum if the first and last terms are 1 and each term differs from its neighbours by at most. For example, $1+2+2+3+3+2+1$ and $1+2+3+4+3+2+1$ are cool sums. The minimum number of terms required to write 2019 as a cool sum is ……
Sequence is
$$
\begin{aligned}
1+2+3+\ldots . .+19+20+21 &+22+23+\ldots \ldots +39+39+40+41+42+43+44+44+43+42+\ldots \ldots+3+2+1
\end{aligned}
$$
So minimum number of terms is $89$.
For each positive integer $n$ let $f(n)=n^{4}-3 n^{2}+9$. Then the sum of all $f(n)$ which are prime is
$f(n)=n^{4}-3 n^{2}+9$
$=n^{4}+6 n^{2}-9 n^{2}+9$
$=\left(n^{2}+3\right)^{2}-(3 n)^{2}$
$=\left(n^{2}+3 n+3\right)\left(n^{2}-3 n+3\right)$
$f(n)$ is prime $\Rightarrow n^{2}-3 n+3=1 \Rightarrow n=1,2$
$f(1)=7, f(2)=13$
sum of values $=7+13=20$
The product of four positive integers $a, b, c$ and $d$ is 9 ! The number $a, b, c, d$ satisfy $a b+a+b=$ 1224, $b c+b+c=549$ and $c d+c+d=351$. The $a+b+c+d=\ldots \ldots$
$$
\text { abcd }=\angle 9=9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1
$$
given $a b+a+b=1224 \quad \Rightarrow(1+a)(1+b)=49 \times 25=1225$
Therefore $b c+b+c=549 \quad \Rightarrow(1+b)(1+c)=25 \times 22=550$
and $c d+c+d=351 \Rightarrow(1+c)(1+d)=352=22 \times 16$
So $1+a=49,1+b=25,1+c=22,1+d=16$
$$
a=48, b=24, c=21, d=15
$$
The number of 6 digit numbers of the form "ABCABC", which are divisible by 13 , where $A, B$ and $C$ are distinct digits, $A$ and $C$ being even digits is
$1001 \times \mathrm{ABC}=\mathrm{ABCABC}$ where $1001=13 \times 7 \times 11$ Now $\mathrm{A}$ and $\mathrm{C}$ are even digits and $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are different digits .
Case-I: When $\mathrm{C}$ is zero
Case-II : When $\mathrm{C}$ is not zero
Total number of $6$ digits
Number possible $= 32 + 96 = 128$
In the sequence $1,4,8,10,16,21,25,30$ and 43 , the number of blocks of consecutive terms whose sums are divisible by 11 is
$4+8+10=22$
$8+10+16+21=55$
$8+10+16+21+25+30=110$
$25+30=55$
Exactly four
In the subtraction below, what is the sum of the digits in the result ? $111 \ldots \ldots \ldots . .111$ (100 digits) $-222 \ldots . . .222$ (50 digits)
$1111 \ldots \ldots \ldots \ldots \ldots \ldots .111 \ldots \ldots \ldots \ldots 111$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 222 \ldots \ldots \ldots \ldots .222$
---------------------------------------------
$1111 \ldots \ldots \ldots \ldots \ldots \ldots .108 \ldots \ldots \ldots \ldots .889$
49 times 1,49 times 8 and 1 times 0 and 9
Sum $=49 \times 1+8 \times 49+9$
Therefore
$\Rightarrow \quad 49+392+9=450 .$
The least odd prime factor of $2019^{8}+1$ is
Let $P$ be an odd prime which divides $2019^{8}+1$
So $2019^{8} \equiv-1(\bmod P)$
$\Rightarrow \quad 2019^{16} \equiv 1(\bmod P)$
Now by Euler's theorem
$2019^{P-1} \equiv 1(\bmod \mathrm{P})$
So $P-1$ should be divisible by $16$
Where $P$ is a prime
First two prime numbers which gives remainder $1$ when divided by $16$ is $17$ and $97$
Case-1 $\quad \mathrm{P}=17$
$2019^{8}+1 \equiv 13^{8}+1 \equiv 4^{8}+1 \equiv 16^{4}+1 \equiv 2 $(mode $17$)
While
$2019^{8}+1=79^{8}+1=18^{8}+1=324^{4}+1=33^{4}+1=1089^{2}+1=22^{2}+1=485=0$ (mod $97$)
So the answer is $97$ .
Let $a, b, c$ be positive integers each less than 50 , such that $a^{2}-b^{2}=100 c$. The number of such triples $(a, b, c)$ is
$a^{2}-b^{2}=100 c$
As $a^{2}-b^{2}$ is a multiple of $100 .$
So it means the last $2$ digit of $a^{2}$ and $b^{2}$ is same.
So $(a, b)$ can be $(49,1)(48,2)(47,3) \ldots \ldots \ldots \ldots .(26,24)$ .
So there are $24$ such pairs .
One more pair for $(a, b)$ is $(25,15)$ .
So total $25$ pairs are possible.
If $4921 \times D=A B B B D$, then the sum of the digits of $A B B B D \times D$ is
4-digit no. (4921) is multiplied by a single digit no. (D) \& result is five digit no., so definitely $D>2$
So by hit \& trial we put the values of D from 3 to 9 .
at $\mathrm{D}=7$
$4921 \times 7=34447$
$(\mathrm{ABBBD})$
So $\mathrm{A}=3, \quad \mathrm{~B}=4, \mathrm{D}=7$
Now ABBBD $(34447) \times 7=241129$
Sum of digits $=2+4+1+1+2+9=19$
What is the $2019^{\text {th }}$ digit to the right of the decimal point, in the decimal representation of $\frac{5}{28}$ ?
$$
\frac{5}{28}=\cdot 17 \overline{857142}
$$
$\Rightarrow \quad 2019=2+336 \times 6+1$ [2 for $17 \& 336$ pairs of 6 repeating number]
$2019^{\text {th }}$ digit from right side to decimal is first digit in repetition
So correct answer is $8$.
If $\mathrm{X}$ is a 1000 digit number, $Y$ is the sum of its digits, $Z$ the sum of the digits of $Y$ and $W$ the sum of the digits of $Z$, then the maximum possible value of $W$ is
$\mathrm{X} \rightarrow 1000$ digit no. If all digit are ' 9 ' so that maximum sum of digit of ' $\mathrm{X}$ ' is 9000 So maximum value of $Y$ is 9000
But for maximum sum of digit of $Y$ is 35 for number (8999) So $Z$ is maximum 35 .
Now for maximum sum of digit of $Z$ is 11 for number $29 .$ So $W=11$. Practical example: if $\quad X=\underbrace{99 \ldots 9}_{333 \text { times }} \quad \underbrace{000 \ldots 0}_{666 \text { times }} 2$ Sum of digit of $X=Y=2999$ Sum of digit of $Y=Z=29$ Sum of digit of $Z=W=11$
Let $x$ be the number $0.000$. 001 which has 2019 zeroes after the decimal point. Then which of the following numbers is the greatest?
(A) $10000+x$, (B) $10000 \cdot x$, (C) $\frac{10000}{x}$, (D) $\frac{1}{x^{2}}$
$$
x=\cdot 000 \ldots 01=10^{-2020}
$$
From option $(A)=10000+x=1000+10^{-2020}=10000 \cdot \underbrace{000 \ldots .01}_{2019 \text { times }}$
From option $(\mathrm{B})=10000 \times \mathrm{x}=10^{4} \times 10^{-2020}=10^{-2016}$
From option $(C)=\frac{10000}{x}=\frac{10^{4}}{10^{-2020}}=10^{2024}$
From option $(D)=\frac{1}{x^{2}}=\frac{1}{\left(10^{-2020}\right)^{2}}=10^{4040}$
So from options $\frac{1}{x^{2}}$ is greatest.
$A B C$
then the number of possible values for $A, B, C, D, E$ satisfying this equation where
If $\frac{C \mathrm{CBA}}{\mathrm{DEDD}}$ then the number of $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ and $\mathrm{E}$ are distinct digits is
$\quad \quad ABC$
$+\quad CBA$
_________
$\quad DEDD$
$D \rightarrow$ must be $1$
Means $C+A=1$ or 11
$So \quad$ Sum of $B+B$ is even $\& D$ is $1 .$
So possible values of B is 0 or 5 .
But if we take $B$ as ' 0 ' so there is no carry forward \& Sum of $A \& C$, did not get different digit from D.
So B must be 5 .
Sum is convert into
$A 5 C$
$C 5 A$
_____
$1211$
Now possible pairs of $(A, C)$ are $(3,8)(4,7)(7,4)(8,3)$
So total 4 possible solutions are there.
$\mathrm{R} A \mathrm{~T}$
If $+\mathrm{MAC}$ and each alphabet represents a different digit, what is the maximum possible value $+\frac{\mathrm{VAT}}{\mathrm{FLAT}}$
of FLAT?
T should be 0 or 5
But if we take $T$ as ' 5 ' sum is $15 \& 1$ carry is forward and sum of 3 'A' and '1' never give unit digit 'A' So T must be '0'
Now again possible values of $A$ are $' 0 ' \& 5$ but alphabet represents different digits so $\mathrm{A}$ is 5 . For maximum value of FLAT, we take maximum value of $R, M \& V$ as $9,8 \& 7$, but sum is $25 \& 5$ is repeat.
So by taking $R, M \& V$ as $9,8 \& 6$. We get maximum value of FLAT.
$$
$\ \ 9 5 0 $
$\ \ 8 5 0 $
$\ \ 6 5 0 $
______
$ 2 4 5 0 $
$FLAT \Rightarrow 2450$
There are exactly 5 prime numbers between 2000 and 2030 . Note: $2021=43 \times 47$ is not a prime number. The difference between the largest and the smallest among these is
Prime no's between $2000 \& 2030$ are $2003,2011,2017,2027,2029$
Difference between $2029 \& 2003$ is
$$
2029-2003=26
$$
A teacher asks 10 of her students to guess her age. They guessed it as $34,38,40,42,46,48,51$, 54,57 and 59 . Teacher said "At least half of you guessed it too low and two of you are off by one. Also my age is a prime number". The teacher's age is
Age of teacher is greater than $46$ .
Again according to questions two of them are off by one.
So there are two possibilities $47 \&\ 58$
$47$ from $(46 \&\ 48)$
$58$ from $(57 \&\ 59)$
But $58$ is not a prime so age of teacher in $47$ .
The sum of 8 positive integers is 22 and their LCM is 9 . The number of integers among these that are less than 4 is
L.C.M. is 9 , so all numbers are from $1,3 \&\ 9$. If we take 2 times 9 sum as 22 is not possible so, 9 will come only one time $\&$ remaining 7 numbers are from $1 \&\ 3$.
So 4 times $1 \&\ 3$ times 3 will come.
$$
1+1+1+1+3+3+3+9=22
$$
So numbers less than 4 is 7
The number of natural numbers $n \leq 2019$ such that $\sqrt[3]{48 n}$ is an integer is
So $\mathrm{n}$ should be multiple of $2^{2} \times 3^{2}=36$ for integer value.
$\mathrm{n} \leq 2019$ $\sqrt[3]{48 \mathrm{n}}=$ integer $\sqrt[3]{2^{4} \times 3 \times n}=$ integer $2 \times \sqrt[3]{2 \times 3 n}=$ integer.
So $\mathrm{n}$ should be multiple of $2^{2} \times 3^{2}=36$ for intege So, possible 'n' are $=2^{2} \times 3^{2}=36<2019$
$=2^{2} \times 3^{2} \times 2^{3}=288<2019$
$=2^{2} \times 3^{2} \times 3^{3}=972<2019$
$=2^{2} \times 3^{2} \times 4^{3}=2304>2019$ (reject)
So only three values of $n$ are possible $36,288,972$
The number of perfect cubes that lie between $2^{9}+1$ and $2^{18}+1$ is
$\left(2^{3}\right)^{3}+1$ and $\left(2^{6}\right)^{3}+1$ $8^{3}+1$ and $64^{3}+1$
Numbers lies between $9,10,11, \ldots \ldots \ldots .64$ Total perfect cubes number $64-8=56$.
