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.

$$\ \ 9 5 0 $$\ \ 8 5 0 $$\ \ 6 5 0 $   ______$ 2 4 5 0 $$FLAT \Rightarrow 2450$     <h3>NMTC 2019 Sub Junior Stage 1 Question 13</h3>   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

$    <h3>NMTC 2019 Sub Junior Stage 1 Question 23</h3>   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 $47$ .     <h3>NMTC 2019 Sub Junior Stage 1 Question 24</h3>   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 $64-8=56$ .