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# NMTC Number Theory Problems and Solutions

### NMTC 2010 Primary Stage 1 Question 1

$\mathrm{n}, \mathrm{a}$ 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 $14$.

### NMTC 2019 Inter Stage 1 Question 17

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

### NMTC 2019 Inter Stage 1 Question 20

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 $89$.

### NMTC 2019 Inter Stage 1 Question 25

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 $=7+13=20$

### NMTC 2019 Inter Stage 1 Question 30

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 $1+a=49,1+b=25,1+c=22,1+d=16$
$$a=48, b=24, c=21, d=15$$

### NMTC 2019 Junior Stage 1 Question 1

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 $= 32 + 96 = 128$

### NMTC 2019 Junior Stage 1 Question 6

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

### NMTC 2019 Junior Stage 1 Question 10

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

$\Rightarrow \quad 49+392+9=450 .$

### NMTC 2019 Junior Stage 1 Question 26

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 $97$ .

### NMTC 2019 Junior Stage 1 Question 27

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.

### NMTC 2019 Sub Junior Stage 1 Question 1

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 $=2+4+1+1+2+9=19$

### NMTC 2019 Primary Stage 1 Question 2

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 $8$.

### NMTC 2019 Sub Junior Stage 1 Question 3

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 $Z=W=11$

### NMTC 2019 Sub Junior Stage 1 Question 4

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.

### NMTC 2019 Sub Junior Stage 1 Question 5

$A B C$
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.

### NMTC 2019 Sub Junior Stage 1 Question 9

$\mathrm{R} A \mathrm{~T}$
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.
2029-2003=26
$$### NMTC 2019 Sub Junior Stage 1 Question 23 A teacher asks 10 of her students to guess her age. They guessed it as 34,38,40,42,46,48,51, 54,57 and 59 . Teacher said "At least half of you guessed it too low and two of you are off by one. Also my age is a prime number". The teacher's age is Age of teacher is greater than 46 . Again according to questions two of them are off by one. So there are two possibilities 47 \&\ 58 47 from (46 \&\ 48) 58 from (57 \&\ 59) But 58 is not a prime so age of teacher in 47 . ### NMTC 2019 Sub Junior Stage 1 Question 24 The sum of 8 positive integers is 22 and their LCM is 9 . The number of integers among these that are less than 4 is L.C.M. is 9 , so all numbers are from 1,3 \&\ 9. If we take 2 times 9 sum as 22 is not possible so, 9 will come only one time \& remaining 7 numbers are from 1 \&\ 3. So 4 times 1 \&\ 3 times 3 will come.$$
1+1+1+1+3+3+3+9=22
$$So numbers less than 4 is 7 ### NMTC 2019 Sub Junior Stage 1 Question 25 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 ### NMTC 2019 Sub Junior Stage 1 Question 30 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. ### NMTC 2010 Primary Stage 1 Question 1 \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. ### NMTC 2019 Inter Stage 1 Question 17 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 ### NMTC 2019 Inter Stage 1 Question 20 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}
$$So minimum number of terms is 89. ### NMTC 2019 Inter Stage 1 Question 25 For each positive integer n let f(n)=n^{4}-3 n^{2}+9. Then the sum of all f(n) which are prime is f(n)=n^{4}-3 n^{2}+9 =n^{4}+6 n^{2}-9 n^{2}+9 =\left(n^{2}+3\right)^{2}-(3 n)^{2} =\left(n^{2}+3 n+3\right)\left(n^{2}-3 n+3\right) f(n) is prime \Rightarrow n^{2}-3 n+3=1 \Rightarrow n=1,2 f(1)=7, f(2)=13 sum of values =7+13=20 ### NMTC 2019 Inter Stage 1 Question 30 The product of four positive integers a, b, c and d is 9 ! The number a, b, c, d satisfy a b+a+b= 1224, b c+b+c=549 and c d+c+d=351. The a+b+c+d=\ldots \ldots$$
\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
$$### NMTC 2019 Junior Stage 1 Question 1 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 ### NMTC 2019 Junior Stage 1 Question 6 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 ### NMTC 2019 Junior Stage 1 Question 10 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 . ### NMTC 2019 Junior Stage 1 Question 26 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 . ### NMTC 2019 Junior Stage 1 Question 27 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. ### NMTC 2019 Sub Junior Stage 1 Question 1 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 ### NMTC 2019 Primary Stage 1 Question 2 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}
$$\Rightarrow \quad 2019=2+336 \times 6+1 [2 for 17 \& 336 pairs of 6 repeating number] 2019^{\text {th }} digit from right side to decimal is first digit in repetition So correct answer is 8. ### NMTC 2019 Sub Junior Stage 1 Question 3 If \mathrm{X} is a 1000 digit number, Y is the sum of its digits, Z the sum of the digits of Y and W the sum of the digits of Z, then the maximum possible value of W is \mathrm{X} \rightarrow 1000 digit no. If all digit are ' 9 ' so that maximum sum of digit of ' \mathrm{X} ' is 9000 So maximum value of Y is 9000 But for maximum sum of digit of Y is 35 for number (8999) So Z is maximum 35 . Now for maximum sum of digit of Z is 11 for number 29 . So W=11. Practical example: if \quad X=\underbrace{99 \ldots 9}_{333 \text { times }} \quad \underbrace{000 \ldots 0}_{666 \text { times }} 2 Sum of digit of X=Y=2999 Sum of digit of Y=Z=29 Sum of digit of Z=W=11 ### NMTC 2019 Sub Junior Stage 1 Question 4 Let x be the number 0.000. 001 which has 2019 zeroes after the decimal point. Then which of the following numbers is the greatest? (A) 10000+x, (B) 10000 \cdot x, (C) \frac{10000}{x}, (D) \frac{1}{x^{2}}$$
x=\cdot 000 \ldots 01=10^{-2020}
$$From option (A)=10000+x=1000+10^{-2020}=10000 \cdot \underbrace{000 \ldots .01}_{2019 \text { times }} From option (\mathrm{B})=10000 \times \mathrm{x}=10^{4} \times 10^{-2020}=10^{-2016} From option (C)=\frac{10000}{x}=\frac{10^{4}}{10^{-2020}}=10^{2024} From option (D)=\frac{1}{x^{2}}=\frac{1}{\left(10^{-2020}\right)^{2}}=10^{4040} So from options \frac{1}{x^{2}} is greatest. ### NMTC 2019 Sub Junior Stage 1 Question 5 A B C then the number of possible values for A, B, C, D, E satisfying this equation where If \frac{C \mathrm{CBA}}{\mathrm{DEDD}} then the number of \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} and \mathrm{E} are distinct digits is \quad \quad ABC +\quad CBA _________ \quad DEDD D \rightarrow must be 1 Means C+A=1 or 11 So \quad Sum of B+B is even \& D is 1 . So possible values of B is 0 or 5 . But if we take B as ' 0 ' so there is no carry forward \& Sum of A \& C, did not get different digit from D. So B must be 5 . Sum is convert into A 5 C C 5 A _____ 1211 Now possible pairs of (A, C) are (3,8)(4,7)(7,4)(8,3) So total 4 possible solutions are there. ### NMTC 2019 Sub Junior Stage 1 Question 9 \mathrm{R} A \mathrm{~T} If +\mathrm{MAC} and each alphabet represents a different digit, what is the maximum possible value +\frac{\mathrm{VAT}}{\mathrm{FLAT}} of FLAT? T should be 0 or 5 But if we take T as ' 5 ' sum is 15 \& 1 carry is forward and sum of 3 'A' and '1' never give unit digit 'A' So T must be '0' Now again possible values of A are ' 0 ' \& 5 but alphabet represents different digits so \mathrm{A} is 5 . For maximum value of FLAT, we take maximum value of R, M \& V as 9,8 \& 7, but sum is 25 \& 5 is repeat. So by taking R, M \& V as 9,8 \& 6. We get maximum value of FLAT.$$

$\ \ 9 5 0$
$\ \ 8 5 0$
$\ \ 6 5 0$

______
$2 4 5 0$

$FLAT \Rightarrow 2450$

### NMTC 2019 Sub Junior Stage 1 Question 13

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
$$2029-2003=26$$

### NMTC 2019 Sub Junior Stage 1 Question 23

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 $47$ .

### NMTC 2019 Sub Junior Stage 1 Question 24

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

### NMTC 2019 Sub Junior Stage 1 Question 25

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 $36,288,972$

### NMTC 2019 Sub Junior Stage 1 Question 30

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 $64-8=56$.

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