Number theory/algebra/geometry solved problems (from NMTC 2025 paper)

 Q1.

The solution of

$$\frac{\sqrt[7]{12 + x}}{x} + \frac{\sqrt[7]{12 + x}}{12} = \frac{64}{3} \left(\sqrt[7]{x}\right)$$

is of the form $\frac{a}{b}$, where $a, b$ are natural numbers with $\gcd(a, b) = 1$; then $b - a$ is equal to:

A) 115 B) 114 C) 113 D) 125

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Solution:
 (x+12)^(1/7)[1/x + 1/12] = 64.x^(1/7)/3
 (x+12)^(1/7)[(x + 12)/12x] = 64.x^(1/7)/3
(x+12)^(8/7) = 12.64.x^(8/3)/3
[(x+12)/x]^(8/7) = 256 = 2^8
=> [(x+12)/x]^(1/7) = 2
=>  (x+12)/x = 2^7 = 128
=> x + 12 = 128x => 127x = 12 => x = 12/127 => a = 12, b = 127 => b - a = 115 = Answer

Q2.

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The value of

$$(52 + 6\sqrt{43})^{3/2} - (52 - 6\sqrt{43})^{3/2}$$

is:

A) 858 B) 918 C) 758 D) 828

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Solution:

Note that:
52 + 6.sqrt(43) = (sqrt(43)  + 3)^2
52 - 6.sqrt(43) = (sqrt(43)  - 3)^2
[52 + 6.sqrt(43)]^(1/2) = (sqrt(43)  + 3)
[52 - 6.sqrt(43)]^(1/2) = (sqrt(43)  - 3)
Let a  = sqrt(43), b = 3
The given expression becomes:
(a + b)^3 - (a - b)^3 = 2b^3 + 2.3.a^2.b = 54 + 6.43.3 = 54 + 774 = 828 = Answer

Q3.

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In the adjoining figure,

$$\angle DCE = 10^\circ, \quad \angle CED = 98^\circ, \quad \angle BDF = 28^\circ$$

Then the measure of angle $x$ is:

A) 72° B) 76° C) 44° D) 82°

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Solution:
Angle EDC = 72, ADC = 108.
ADCB is a cyclic quadrilateral, hence Angle ABC = 180 - ADC = 180 - 108 = 72.
GBA = 180 - ABC = 108
The chord BF subtends angle BDF = 28 so BAF = 28 as well.
In triangle AGB, x + ABG + BAG = 180 = x + 108 + 28 => x = 180 - 136 = 44 = Answer.

Q4.

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The sum of all positive integers $m, n$ which satisfy

$$m^2 + 2mn + n = 44$$

is _______

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Solution:
n(2m + 1) = 44 - m^2
=> n = (44 - m^2)/(2m + 1)
Since n is positive integer, m^2 < 44.
=> m can be 1,2,3,4,5,6
m = 1 => n = 43/3 not integer
m = 2 => n = 40/5 = 8 is integer
m = 3 => n = 35/7 = 5 is integer
m = 4 => n = 28/8 not integer
m = 5 => n = 19/11 not integer
m = 6 => 8/13 not integer
Valid (m,n) pairs are (2,8) and (3,5).
Add them to get 17 = answer.

Q5. 

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The expression

$$49(a + b)^2 - 46(a - b)^2$$

is factorized into

$$(la + mb)(na + pb),$$

then the numerical value of $l + m + n + p$ is ______.

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It can be written as:
[7(a + b)]^2 - [sqrt(46).(a-b)]^2
= [7a + 7b + sqrt(46).a - sqrt(46).b][7a + 7b - sqrt(46).a + sqrt(46).b]
=> l = 7 + sqrt(46), m = 7 - sqrt(46), n = 7 - sqrt(46), p = 7 + sqrt(46)
Add all to get:
28 = answer

Q6.

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The number of solutions $(x, y)$ of the simultaneous equations

$$\log_4 x - \log_2 y = 0, \quad x^2 = 8 + 2y^2$$

is _______.

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Solution:
First equation becomes sqrt(x)/y = 1 => x = y^2
Substitute in second equation and solve.
x^2 = 8 + 2x => x^2 - 2x - 8 = 0 => (x-4)(x+2) = 0 => x = -2,4
But x > 0 else log can't be taken.
=> x = 4, y = -+2
Again y = 2 else log is undefined.
So only pair of (x,y) is (4,2). Answer = 1.

Q7.

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 $a, b, c$ are real numbers such that

$$b - c = 8 \quad \text{and} \quad bc + a^2 + 16 = 0.$$

The numerical value of

$$a^{2025} + b^{2025} + c^{2025}$$

is _______.

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Solution:
b = 8 + c substitute in second equation:
(8 + c).c + a^ + 16 = 0
8c + c^2 + a^2 + 16 = 0 = (c + 4)^2 + a^2 => c = -4, a = 0 => b = 4
So the given expression evaluates to 0.

Q8.

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Given

$$f(x) = \frac{2025x}{x + 1} \quad \text{where } x \ne -1.$$

Then the value of $x$ for which

$$f(f(x)) = (2025)^2$$

is _______.

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Solution:
f(f(x)) = [2025.2025.x/(x+1)]/[2025x/(x + 1) + 1] = 2025^2
=> x/(x+1) = (2025x + x + 1)/(x+1)
=> x = 2026x + 1 => x = -1/2025

Q9.

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The number of integral solutions of the inequation

$$\left| \frac{2}{x - 13} \right| > \frac{8}{9}$$

is:

A) 1  B) 2  C) 3  D) 4

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Solution:
Invert to get:
|(x-13)/2| < 9/8 (x != 13)
-9/8 < (x-13)/2 < 9/8
=> -9/4 < x-13 < 9/4
=> 43/4 < x < 61/4
But x is integer => 
11 <= x <= 15
So x = 11,12,14,15(it can't be 13). So 4 solutions.

Q10.

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$$\frac{\sqrt{a + 3b} + \sqrt{a - 3b}}{\sqrt{a + 3b} - \sqrt{a - 3b}} = x,$$

then the value of

$$\frac{3bx^2 + 3b}{ax}$$

is:

A) 1  B) 2  C) 3

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Solution:
The given expression can be written as:
(3b/a)[(x^2 + 1)/x] = (3b/a)[x + 1/x]
WLOG, a = 4, b = 1
x = (sqrt(7) + 1)/(sqrt(7) - 1)
x + 1/x = (sqrt(7) + 1)/(sqrt(7) - 1) + (sqrt(7) - 1)/(sqrt(7) + 1)
 = 16/6 = 8/3
3b/a = 3/4
Finally: (3/4) * (8/3) = 2 = Answer

Q11.

 $ABC$ is a right triangle in which $\angle B = 90^\circ$.
The inradius of the triangle is $r$ and the circumradius of the triangle is $R$.

If $R : r = 5 : 2$, then the value of

$$\cot^2\left(\frac{A}{2}\right) + \cot^2\left(\frac{C}{2}\right)$$

is:

A) $\frac{25}{4}$  B) 17  C) 13  D) 14

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Solution:
Prerequisites:
Right angle inradius and circumradius.
Cotangent half angle identity.

For a right triangle:
Circumradius R = h/2 where h is hypotenuse.
Inradius r = (x + y - h)/2 where x,y are the legs.
R/r = 5/2 => h = 5r
x + y - 5r = 2r => x + y = 7r
x^2 + y^2 = h^2 = 25r^2
WLOG r = 1 => x + y = 7 and x^2 + y^2 = 25, a classic 3-4-5 right triangle.
We know that
cot(A/2) = (1 + cosA)/sinA
WLOG sinA = 3/5 and cosA = 4/5
cot(A/2) = 9/3 = 3
cot(C/2) = (1 + cosC)/sinC
But cosC = sinA and sinC = cosA
So cot(C/2) = (1 + 3/5)/(4/5) = 2
3^2 + 2^2 = 13 = Answer