Mock test 5 - pending 21,23-28

Q1. (Combinatorics)

"How many ordered quadruples (a, b, c, d) of positive odd integers are there that satisfy the equation a + b + c + 2d = 15?"

Solution:

Answer: 34

Why

1. Rewrite every odd variable in terms of non-negative integers

$$a = 2x+1,\; b = 2y+1,\; c = 2z+1,\; d = 2w+1,\qquad x,y,z,w \ge 0$$

2. Plug into the equation

$$(2x+1)+(2y+1)+(2z+1)+2(2w+1)=15 \Rightarrow 2x+2y+2z+4w+5 = 15 \Rightarrow 2(x+y+z+2w)=10 \Rightarrow x+y+z+2w = 5$$

3. Count solutions for $x+y+z+2w=5$

Because $w$ is an integer, list its possible values and use stars-and-bars for the remaining sum:

$w$	equation for $x+y+z$	# solutions $\binom{n+2}{2}$
0	$x+y+z = 5$	$\binom{7}{2}=21$
1	$x+y+z = 3$	$\binom{5}{2}=10$
2	$x+y+z = 1$	$\binom{3}{2}=3$

Higher $w$ makes $2w>5$, so there are no more cases.

4. Add them up

$$21 + 10 + 3 = 34$$

Each quadruple $(x,y,z,w)$ corresponds uniquely to a quadruple of positive odd integers $(a,b,c,d)$. Hence, 34 ordered quadruples satisfy $a+b+c+2d = 15$.

Q2. (algebra)

"Find (x + 2y + 3z) from the system of equations xy = 27, xz = 8 and yz = 6."

Solution:
It's straightforward variable replacement and solving.
You will get z = -+4/3 x = -+6 y = -+9/2.
Now evaluating the given expression will give -+19.
Method 2:
Multiply all and take square root to get xyz = 36
Then divide each equation to get individual values.

Q3. (number-theory)

"I am thinking of two numbers between 1 and 20. The first is one less than a perfect square and is not prime. The second is one more than a perfect square and is prime. The sum of the two numbers is divisible by 4 but not by 8. What is the product of the two numbers?"

Solution:
Let the numbers be a,b.
a = 8, 15
b = 5, 17
a + b = 15 + 5 = 20 is the only sum satisfying the given condition.
So answer : 15 * 5 = 75.

Q4. (number-theory)

"Find the positive integer x such that x(x + 1)(x + 2)(x + 3) + 1 = 271^2."

Solution:
 x(x + 1)(x + 2)(x + 3) = 271^2 - 1 = 272*270
272 = 16*17
270 = 15*18
=> x = 15

Q5. (algebra)

Three positive numbers are given such that:
(i) The first of the numbers is half the second.
(ii) The product of the first and the second number is equal to the sum of the second and the third number.
(iii) The third number is three times as large as the second.

Find the first of the given numbers.

Solution:
a,b,c
a = b/2
ab = b + c
c = 3b
=> ab = 4b => a = 4 = answer

Q6. (number-theory,logarithm)

"How many real solutions x are there to the following equation:

log₂(x⁴ + 2x³ + 5x² + 6x + 1) = 3·log₂(x + 1) + 1."

Solution:

Answer: 1 real solution

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Step-by-step reasoning

1. Rewrite the right–hand side using log rules

$$3\log_2(x+1)+1 \;=\; \log_2\!\bigl((x+1)^3\bigr)+\log_2 2 \;=\; \log_2\!\bigl(2\,(x+1)^3\bigr).$$

2. Drop the logarithm (arguments must stay positive)

Because $\log_2 A = \log_2 B$ implies $A=B>0$,

$$x^{4}+2x^{3}+5x^{2}+6x+1 \;=\; 2\,(x+1)^3 .$$

3. Expand and simplify

$$x^{4}+2x^{3}+5x^{2}+6x+1 -\;2\bigl(x^{3}+3x^{2}+3x+1\bigr) =\; x^{4}-x^{2}-1 =0.$$

4. Solve the quartic via a quadratic in $y=x^{2}$

$$y^{2}-y-1=0 \;\Longrightarrow\; y=\frac{1\pm\sqrt{5}}{2}.$$

Only $y=\dfrac{1+\sqrt5}{2}>0$ is admissible, so

$$x=\pm\sqrt{\frac{1+\sqrt5}{2}}.$$

5. Check the domain constraints

• For the logarithms we need $x+1>0\;\Longrightarrow\;x>-1$.

• The positive root $x=\sqrt{\tfrac{1+\sqrt5}{2}}\approx1.272$ satisfies $x>-1$.

• The negative root $x=-\sqrt{\tfrac{1+\sqrt5}{2}}\approx-1.272$ does not (since $x+1<0$), so it is excluded.

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Conclusion

Only the positive root survives the domain check, so exactly one real solution exists.

Q7. 

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In triangle ABC, D and E are points on sides AB and AC respectively, such that BE is perpendicular to CD. Let X be a point inside the triangle such that
$\angle XBC = \angle EBA$ and $\angle XCB = \angle DCA$. If $\angle A = 54^\circ$, what is the measure of $\angle EXD$?

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Solution 1:

Concise solution:

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Knowing the answer is numerical, we take the limit as $D$ approaches $A$. Then $E$ is the foot from $B$ to $AC$, and $X$ coincides with $B$. Then $\angle EXD = \angle EBA$. But triangle $BEA$ is right, hence
$\angle EBA = 90^\circ - \angle EAB = 36^\circ$.

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Detailed solution:
What we are trying to do here is this:
The question doesn't give fixed positions of D,E,X. We will move them in a way that original conditions are satisfied and we get a solution.
Let's say we take D very close to A. Given that angles XCB = ACD the ray CX will almost coincide with CB and XCB = ACD = very small positive angle.
Since BE is perpendicular to CD, BE will almost become a perpendicular to AC and EBA is almost a right triangle with right angle at E. So EBA = 90 - 54 = 36 almost.
Since EBA = 36 so XBC = 36 as well.
XB and XC meet at X and X will be very close to B such that it still maintains XBC = 36 angle.
Why is X close to B?
Because as shown earlier X is very close to the line BC. If X is away from B towards 'C' then XBC will become close to 0 and not 36 which we need.
So now we have established that X is close to B, D close to A and E is on AC s.t. BE is perpendicular to AC.
So EXD ~= EBA = 36 = Answer.

Same solution in much more detail:

What the “$D\to A$” trick is really doing

The official solution first observes that the numerical value of $\angle EXD$ does not depend on where $D$ (and therefore $E$ and $X$) sit on the sides as long as the given angle-equalities are satisfied.
Because of that, we can slide $D$ along $AB$ to any convenient position, or even let it approach the vertex $A$. The equal-angle conditions keep producing a (unique) point $X$ for every admissible $D$. When we “take the limit $D\to A$” we look at what happens to all the other objects continuously; we are not claiming that the configuration with $D=A$ itself still satisfies the original conditions.

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How $E$ and $X$ move when $D$ is very close to $A$

Object	Behaviour as $D\to A$	Reason
$E$	Slides to the foot of the perpendicular from $B$ to $AC$	$BE\perp CD$, and $CD$ collapses onto $AC$
$\angle EBA$	Tends to a fixed value $90^\circ-\angle A\;=\;36^\circ$	Triangle $BEA$ becomes right-angled at $E$
$X$	Slides along a path that converges to $B$	See explanation below

Why $X\to B$.
Let

$$\alpha=\angle DCA\quad\text{and}\quad \beta=\angle EBA .$$

• As $D\to A$ we have $\alpha\to 0$.
  The condition $\angle XCB=\alpha$ forces the ray $CX$ to hug the side $CB$ ever more tightly.

• Meanwhile $\beta\to 36^\circ$.
  The other condition $\angle XBC=\beta$ keeps $X$ on the fixed ray from $B$ that makes $36^\circ$ with $BC$.

For very small $\alpha$ those two rays—one starting at $C$ almost coinciding with $CB$, the other starting at $B$ tilted $36^\circ$ inside the triangle—meet extremely close to $B$. Hence $X$ drifts toward $B$ (and no other point).

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What about the angle conditions at the exact limit?

Exactly at $D=A$ the original equal-angle requirements become degenerate:

• $\angle DCA$ is $0^\circ$, so we can no longer ask for a positive angle $\angle XCB$.

• $\angle XBC=\angle EBA$ cannot both be satisfied if $X=B$.

That is not a contradiction—the limiting configuration is not meant to satisfy the problem’s hypotheses.
It is only a device that lets us read off the desired angle, because that angle varies continuously as the points move.

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Computing the limit of $\angle EXD$

When $D$ is extremely close to $A$ and $X$ is correspondingly very close to $B$,

$$\angle EXD\;\approx\;\angle (\;XB,\;XD\;)\;=\;\angle (\;BE,\;BA\;)=\angle EBA .$$

Taking the exact limit gives

$$\boxed{\displaystyle \lim_{D\to A}\angle EXD=\angle EBA } .$$

Finally, in the right-angled triangle $BEA$,

$$\angle EBA=90^\circ-\angle EAB=90^\circ-\angle A=90^\circ-54^\circ=36^\circ .$$

So $\angle EXD=36^\circ$.

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Key take-away

“$X$ coincides with $B$” is shorthand for “$X$ approaches $B$ in the limiting configuration we are using to evaluate the desired angle.”
The original angle conditions are applied before taking the limit; afterwards we only keep the angle $\angle EXD$, whose limit is well-defined and equals $36^\circ$.

Solution 2:(not gone through yet)

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Observe that as $\angle XBZ = \angle DBY$ and $\angle DYB = \angle XZB = 90^\circ$, triangles $DBY$ and $XBZ$ are similar by AA. Then

$$\frac{DB}{XB} = \frac{YB}{ZB}, \text{ or } \frac{DB}{YB} = \frac{XB}{ZB}.$$

Combined with $\angle DBY = \angle XBZ$ again, we get triangles $DBX$ and $YBZ$ are similar by SAS. Similarly, triangles $ECX$ and $YCZ$ are similar.

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Using quadrilateral $ABYC$, and then triangle $XBC$, we get:

$$\begin{aligned} 360^\circ &= \angle BAC + \angle ACY + \angle CYB + \angle YBA \\ 360^\circ &= 54^\circ + \angle ACY + 270^\circ + \angle YBA \\ 36^\circ &= \angle ACY + \angle YBA \\ 36^\circ &= \angle XCB + \angle XBC \\ 144^\circ &= \angle BXC \end{aligned}$$

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Also, by the two similarities,

$$\angle DXB + \angle EXC = \angle YZB + \angle YZC = 180^\circ.$$

Thus, we get

$$\angle DXE = 360^\circ - \angle DXB + \angle EXC - \angle BXC = 360^\circ - 180^\circ - 144^\circ = 36^\circ.$$

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Q8(algebra). Find xyz if x, y, and z are positive reals satisfying the system.

$$x + \frac{1}{y} = 4,\quad y + \frac{1}{z} = 1,\quad z + \frac{1}{x} = \frac{7}{3}$$

Solution:
Just keep replacing variables and solve the quadratic, you will get:
y = 2/5, z = 5/3, x = 3/2
Multiply all to get answer = 1.

Method 2:

We want to find the product $xyz$, therefore, we multiply the three equations:

$$\left( x + \frac{1}{y} \right) \left( y + \frac{1}{z} \right) \left( z + \frac{1}{x} \right) = xyz + (x + y + z) + \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right) + \frac{1}{xyz}$$ $$= 4 \cdot 1 \cdot \frac{7}{3} = \frac{28}{3}$$

Adding the equations,

$$\left( x + \frac{1}{y} \right) + \left( y + \frac{1}{z} \right) + \left( z + \frac{1}{x} \right) = 4 + 1 + \frac{7}{3} = \frac{22}{3}$$

Therefore,

$$xyz + \frac{22}{3} + \frac{1}{xyz} = \frac{28}{3} \Rightarrow xyz + \frac{1}{xyz} = 2 \Rightarrow \boxed{xyz = 1}$$

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Q9(number-theory). XYZ is a three-digit number. Given that

XXXX + YYYY + ZZZZ = YXXXZ,
what is X + Y + Z?

Solution:

Solution outline

1. Translate the repeating-digit notation into algebra

• XXXX = 1111 × X

• YYYY = 1111 × Y

• ZZZZ = 1111 × Z

• YXXXZ = 10000 Y + 1110 X + Z

2. Set up the equation

$$1111X + 1111Y + 1111Z \;=\; 10000Y + 1110X + Z$$

3. Rearrange

$$(1111X - 1110X) + (1111Z - Z) \;=\; 10000Y - 1111Y$$ $$\boxed{X + 1110Z = 8889Y}$$

4. Note that all variables are digits (0–9) and $X,\,Y\neq0$ because $XYZ$ is a three-digit number and $Y$ is the leading digit of the 5-digit result.

5. Observe the size constraint:
   Left-hand side ≤ $9 + 1110\times9 = 9999$.
   Right-hand side = $8889Y$.
   Therefore $8889Y \le 9999 \implies Y = 1$.

6. Substitute $Y = 1$:

$$X + 1110Z = 8889$$

Because $X$ is a single digit, $8889 - X$ must be divisible by 1110.

Checking $X = 1,2,\dots,9$:

$$8889 - 9 = 8880 = 1110 \times 8$$

Only $X = 9$ gives an integer $Z = 8$.

7. Verify:

$$9999 + 1111 + 8888 = 19998 = \text{YXXXZ} = 1\,999\,8$$

8. Compute the requested sum:

$$X+Y+Z=9+1+8=\boxed{{\displaystyle 18}}$$

Q10. (algebra,A.P.)

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Suppose that $\{a_n\}_{n \ge 1}$ is an arithmetic sequence of real numbers such that

$$a_1 + a_2 + a_3 + a_4 + \cdots + a_{10} = 20,$$ $$a_1 + a_4 + a_9 + a_{16} + \cdots + a_{100} = 18.$$

Compute $a_1 + a_8 + a_{27} + a_{64} + \cdots + a_{1000}.$

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Solution:

Answer: 2

Why that’s the value

1. Describe the arithmetic sequence

   $$a_n \;=\; a_1 + (n-1)d,$$

   where $a_1$ is the first term and $d$ is the common difference.

2. Use the sum of the first 10 terms

   $$S_{10}=10\cdot\frac{a_1+a_{10}}{2} \;=\;10\cdot\frac{a_1 + \bigl[a_1+9d\bigr]}{2} \;=\;5(2a_1+9d)=20 \;\Longrightarrow\; 2a_1+9d=4.\tag{1}$$

3. Use the sum of the terms whose indices are perfect squares

   Indices: $1^2,4^2,9^2,\dots,10^2$.

   $$\sum_{k=1}^{10} a_{k^2} =\sum_{k=1}^{10}\Bigl[a_1+(k^2-1)d\Bigr] =10a_1+d\sum_{k=1}^{10}(k^2-1).$$

   Since $\sum_{k=1}^{10}k^2=385$ we get

   $$10a_1 + d(385-10)=10a_1+375d = 18.\tag{2}$$

4. Solve the two equations

   From (1): $a_1=\dfrac{4-9d}{2}$.

   Substitute into (2):

   $$10\Bigl(\tfrac{4-9d}{2}\Bigr)+375d=18 \;\Longrightarrow\; 20-45d+375d=18 \;\Longrightarrow\; 330d=-2 \;\Longrightarrow\; d=-\frac{1}{165}.$$

   Then $a_1=\dfrac{4-9(-1/165)}{2}=\dfrac{223}{110}$.

5. Sum the terms whose indices are perfect cubes

   Indices: $1^3,2^3,3^3,\dots,10^3$ (i.e., $1,8,27,\dots,1000$).

   $$\sum_{k=1}^{10} a_{k^3} =\sum_{k=1}^{10}\Bigl[a_1+(k^3-1)d\Bigr] =10a_1+d\bigl(\sum_{k=1}^{10}k^3-10\bigr).$$

   Because $\sum_{k=1}^{10}k^3=(1+2+\dots+10)^2=55^2=3025$,

   $$10a_1 + d(3025-10) = 10\Bigl(\tfrac{223}{110}\Bigr) + \Bigl(-\tfrac{1}{165}\Bigr)\!(3015) = \frac{223}{11} - \frac{201}{11} = \frac{22}{11} = 2.$$

Hence

$$\boxed{{\displaystyle 2}}\mathrm{.}$$

Q11.(number-theory)
How many pairs of positive integers (a, b) are there, both not exceeding 10, such that a ≤ gcd(a, b) ≤ b and a ≤ lcm(a, b) ≤ b?

Solution:

Answer: 27 pairs

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Why the two inequalities force a to divide b

1. gcd condition

   $$a \;\le\; \gcd(a,b) \;\le\; b$$

   • But $\gcd(a,b)\le \min(a,b)$.

   • If we also have $\gcd(a,b)\ge a$, then necessarily $\gcd(a,b)=a$.

   • A greatest common divisor equal to $a$ means $a\mid b$.

   • We’re also allowed to assume $a\le b$ (otherwise $\gcd(a,b)=a$ would be impossible).

2. lcm condition

   $$a \;\le\; \operatorname{lcm}(a,b) \;\le\; b$$

   • For any two positive integers, $\operatorname{lcm}(a,b)\ge \max(a,b)=b$ since $b\ge a$.

   • The upper bound $\operatorname{lcm}(a,b)\le b$ therefore forces $\operatorname{lcm}(a,b)=b$.

3. Compatibility check
   For numbers with $a\mid b$,

   $$\gcd(a,b)=a,\qquad \operatorname{lcm}(a,b)=b,$$

   which satisfies both inequalities automatically.

Hence the problem is equivalent to:

Count ordered pairs $(a,b)$ with $1\le a\le b\le 10$ such that $a\mid b$.

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Counting the divisor pairs

For each $b$ (1 through 10) list its positive divisors $a\le b$.

$b$	Divisors $a$	# pairs
1	1	1
2	1, 2	2
3	1, 3	2
4	1, 2, 4	3
5	1, 5	2
6	1, 2, 3, 6	4
7	1, 7	2
8	1, 2, 4, 8	4
9	1, 3, 9	3
10	1, 2, 5, 10	4
Total		27

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Therefore, there are 27 ordered pairs $(a,b)$ with both numbers not exceeding 10 that meet the given gcd and lcm conditions.

Q12. 

"The diagram, which is not drawn to scale, shows a rectangle divided by a horizontal and a vertical line into four rectangles. The areas of three of them are shown. What is the area of the whole rectangle?"

Solution:

Answer: 35

Why it works

	Left strip (width =$w_1$)	Right strip (width =$w_2$)
Top band (height =$h_1$)	$w_1h_1 = 6$	$w_2h_1 = 9$
Bottom band (height =$h_2$)	$w_1h_2 = 8$	$w_2h_2 = \;?$

1. From the known areas

$$w_1h_1 = 6,\quad w_2h_1 = 9,\quad w_1h_2 = 8$$

2. Solve for the missing corner:

$$w_2h_2 = \frac{(w_2h_1)(w_1h_2)}{w_1h_1} = \frac{9 \times 8}{6} = 12$$

3. Add all four parts to get the whole rectangle’s area:

$$6 + 9 + 8 + 12 = 35$$

So the entire rectangle covers 35 square units.

Q13.(combinatorics)

"The number of positive integers less than 300000 are there that have the digits 5, 6 and 7 next to each other and in that order is M, then sum of digits of M is"

Solution:

Step 1 – Understand the size of the search

All numbers we care about satisfy

$$1 \le n < 300\,000$$

so they have at most 6 digits.

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Step 2 – Treat “567” as one solid block

Because the three digits must appear consecutively and in that order, think of the string 567 as a single “super‐digit” that slides inside the number.
For each total length (3-, 4-, 5-, 6-digit numbers) count the possible positions of that block and then fill the remaining places with any digits allowed by the leading-zero and < 300 000 rules.

Total length	Possible positions of block	Count per position	Sub-total
3	(567)	exactly 1 number → 567	1
4	567x (x = 0–9) → 10x567 (x = 1–9) → 9	10 + 9	19
5	567xy (x,y = 0–9) → 100x567y (x = 1–9, y = 0–9) → 90xy567 (x = 1–9, y = 0–9) → 90	100 + 90 + 90	280
6	• 567abc is invalid (starts with 5 > 2, so ≥ 500 000).• a567bc (a = 1,2; b,c = 0–9) → 2·10·10 = 200• ab567c (a = 1,2; b,c = 0–9) → 200• abc567 (a = 1,2; b,c = 0–9) → 200	0 + 200 + 200 + 200	600

Adding everything:

$$M = 1 + 19 + 280 + 600 = 900.$$

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Step 3 – Find the asked‐for quantity

The problem asks for the sum of the digits of $M$:

$$900 \; \longrightarrow \; 9 + 0 + 0 = 9.$$

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$$\boxed{{\displaystyle 9}}$$

Q14. (geometry)

"ABCD is a trapezium with AB parallel to DC, AB = 4 cm, DC = 11 cm and the area of triangle ABP is 12 cm². What is the area of the trapezium ABCD in square centimetres?"

Solution:

Solution outline

1. Find the height of the trapezium.
   Triangle ABP has base AB = 4 cm and area 12 cm².

$$\text{Area}_{\triangle ABP}= \tfrac12 \times \text{base} \times \text{height} \quad\Longrightarrow\quad 12 = \tfrac12 \times 4 \times h \quad\Longrightarrow\quad h = 6\text{ cm}$$

Since $AB\parallel DC$ and $P$ lies on $DC$, this height $h$ is exactly the perpendicular distance between the two parallel sides—i.e. the height of trapezium $ABCD$.

2. Apply the trapezium area formula.

$$\text{Area}_{ABCD} = \tfrac12 \times (\text{sum of parallel sides}) \times \text{height} = \tfrac12 \times (AB + DC) \times h = \tfrac12 \times (4 + 11)\times 6 = \tfrac12 \times 15 \times 6 = 45\text{ cm}^2$$

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$$\boxed{{\displaystyle 45{\text{ cm}}^2}}$$

Q15. (geometry)

"In a right-angled triangle ABC, a point M on the hypotenuse BC is such that AM ⟂ BC. Also, MC is 8 cm longer than BM, and the ratio AB : AC = 3 : 5. How many centimetres is the length of the hypotenuse?"

Solution:
BC = ?
Note that angles BCA = MCA = MAB and MBA = MAC.
BC = BC + MC
MC = 8 + BM
tan(C) = AB/AC = 3/5 = BM/AM = AM/MC
BM = 3AM/5
MC = 5AM/3 = 8 + 3AM/5 => AM = 15/2
=> BM = 9/2, MC = 25/2
=> BC = 9/2 + 25/2 = 34/2 = 17 = answer

Q16. (number-theory)

Solution:
3 * (9 + 4) * (3^4 - 4^3) + 3 = 3 * 13 * 17 + 3 = 666 = N => N - 600 = 66 = answer

Q17.(number-theory)

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Let the number of positive integers $a \leq 2023$, such that $a$ is even and $\gcd(a, 2023) = 1$ be $N$. Find the sum of digits of $N$.

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Solution:

Solution outline

1. Prime-factorise 2023

$$2023 = 7 \times 17^2$$

So the only prime divisors of 2023 are $7$ and $17$.

2. Translate the condition

An even integer $a\le 2023$ is coprime to 2023 iff it is not divisible by $7$ and not divisible by $17$.

3. Count even numbers up to 2023

$$\bigl\lfloor\tfrac{2023}{2}\bigr\rfloor = 1011$$

4. Use inclusion–exclusion

• Divisible by $2\cdot 7 = 14$:
  $\bigl\lfloor\tfrac{2023}{14}\bigr\rfloor = 144$

• Divisible by $2\cdot 17 = 34$:
  $\bigl\lfloor\tfrac{2023}{34}\bigr\rfloor = 59$

• Divisible by both $14$ and $34$, i.e. by $\operatorname{lcm}(14,34)=238$:
  $\bigl\lfloor\tfrac{2023}{238}\bigr\rfloor = 8$

Applying inclusion–exclusion:

$$N = 1011 \;-\; 144 \;-\; 59 \;+\; 8 \;=\; 816$$

5. Sum of the digits of $N$

$$8 + 1 + 6 = 15$$

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$$\boxed{{\displaystyle 15}}$$

Q18. (algebra)

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"If x and y are real numbers such that $x + y + \sqrt{x + y} = 30$, $x - y + \sqrt{x - y} = 12$. Then $x + y =$"

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Solution:

Let

$$a = x + y, \qquad b = x - y.$$

The two given equations become

$$\begin{cases} a + \sqrt{a} = 30,\\[4pt] b + \sqrt{b} = 12. \end{cases}$$

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1. Solve for $a = x + y$

Set $\sqrt{a}=t\;\;(t\ge 0)$.
Then $a=t^{2}$ and

$$t^{2}+t-30=0 \quad\Longrightarrow\quad t=\frac{-1\pm\sqrt{1+120}}{2} =\frac{-1\pm 11}{2}.$$

The non-negative root is $t=5$, so

$$a=t^{2}=5^{2}=25.$$

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2. (Check consistency)

Likewise, for $b$ set $\sqrt{b}=s\ge 0$:

$$s^{2}+s-12=0 \;\Rightarrow\; s=\frac{-1\pm\sqrt{49}}{2} =\frac{-1\pm 7}{2} \Longrightarrow s=3,\; b=s^{2}=9.$$

Then

$$x=\frac{a+b}{2}=17, \qquad y=\frac{a-b}{2}=8,$$

which indeed satisfy the originals. Hence the solution is consistent.

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$$\boxed{{\displaystyle x+y=25}}$$

Q19. (algebra)
Two years ago, Gene was nine times as old as Carol. He is now seven times as old as she is. How many years from now will Gene be five times as old as Carol?
Solution:
g - 2 = 9(c-2)
g = 7c
g + x = 5(c+x)
Solving, we get:
g = 56
c = 8
x = 4 = answer

Q20. (geometry)

 ABC is an example of an isosceles (but not equilateral) triangle with integer side lengths.

How many such triangles can be made using the integers from 1 to 9 inclusive?

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Solution:
Let the sides be a,a,b. And a != b.
Then a + a > b => 2a > b
a = 1 has only possible b = 1. But a,b can't be same so no solution.
a = 2 =>
2,2,1
2,2,3
2 solutions.
a = 3 => b = 1,2,4,5 => 4 solutions
a = 4 => b = 1,2,3,5,6,7 => 6 solutions
a = 5 => b = 1 to 9 (except 5) => 8 solutions
Same for a = 6,7,8,9 => 32 solutions.
Total = 52 solutions. answer.

Q22. (geometry)

In the figure, 

$B$ is the mid-point of $AD$, $C$ is the mid-point of $DE$, $A$ is the mid-point of $EF$, and $M$ is the midpoint of $AF$. If the area of $\triangle AMB$ is $6 \,\text{cm}^2$, how many $\text{cm}^2$ is the remaining area of $\triangle DEF$?

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Solution:
Triangle AMB is similar to AFD since all sides are parallel. BM = DF/2 by midpoint theorem.
So [AMB] = [AFD]/4 => [AFD] = 24
Also [AFD] = [ADE] because height from D will be same and bases are equal.
So [DEF] = 2*[AFD] = 48
[DEF] - [AMB] = 48 - 6 = 42 = answer.

Q23. (algebra)

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"Ann is four times as old as Mary was when Ann was as old as Mary is now. Furthermore, Ann is twice as old as Mary was when Ann was six years older than Mary is now. How old is Ann?"

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Solution:

Let’s translate the story into algebra

Symbol	Meaning today
A	Ann’s present age (what we want)
M	Mary’s present age

Because all of the statements compare today’s ages with ages in the past, we’ll measure everything from “today” backward.

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1. Re-stating the first sentence

“Ann is four times as old as Mary was when Ann was as old as Mary is now.”

1. Find the moment being described.

   • Today Ann is A and Mary is M.

   • We look back to the moment when Ann’s age equaled Mary’s current age M.

   • Ann must go back A − M years to be that age.

2. Mary’s age at that same moment.

   • She is younger by the same amount of time:
     $M_{\text{then}} = M - (A - M) = 2M - A$.

3. Turn the English into an equation.

   • “Ann (today) is four times Mary (then)”:

     $$A = 4\,(2M - A). \tag{1}$$

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2. Re-stating the second sentence

“Ann is twice as old as Mary was when Ann was six years older than Mary is now.”

1. Find that second moment in the past.

   • We need the time when Ann’s age = M + 6.

   • That is $A - (M + 6)$ years ago.

2. Mary’s age at that moment.

   • $M_{\text{then}} = M - [A - (M + 6)] = 2M - A + 6$.

3. Translate to an equation.

   • “Ann (today) is twice Mary (then)”:

     $$A = 2\,(2M - A + 6). \tag{2}$$

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3. Solve the two equations

From equation (1)

$$A = 8M - 4A \quad\Longrightarrow\quad 5A = 8M \quad\Longrightarrow\quad M = \tfrac{5}{8}A. \tag{3}$$

Substitute $M$ into equation (2)

$$\begin{aligned} A &= 2\!\left[\,2\!\left(\tfrac{5}{8}A\right) - A + 6\right] \\ &= 2\!\left(\tfrac{5}{4}A - A + 6\right) = 2\!\left(\tfrac{1}{4}A + 6\right) = \tfrac{1}{2}A + 12. \end{aligned}$$

Move terms:

$$A - \tfrac{1}{2}A = 12 \;\;\Longrightarrow\;\; \tfrac{1}{2}A = 12 \;\;\Longrightarrow\;\; \boxed{A = 24}.$$

Then from (3):

$$M = \tfrac{5}{8}\times24 = \boxed{15}.$$

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4. Quick check

• First sentence:

  • Moment 1 was $A-M = 9$ years ago.

  • Mary was $15 - 9 = 6$.

  • Ann today is $24$, and $24 = 4 \times 6$. ✔️

• Second sentence:

  • Moment 2 was $A - (M+6) = 24 - 21 = 3$ years ago.

  • Mary was $15 - 3 = 12$.

  • Ann today is $24$, and $24 = 2 \times 12$. ✔️

Both conditions hold, so Ann is 24 years old (and Mary is 15).

Q29.

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Let $f_n(x) = n + x^2$.
If the product

$$\gcd\{f_{2001}(2002), f_{2001}(2003)\} \times \gcd\{f_{2011}(2012), f_{2011}(2013)\} \times \gcd\{f_{2021}(2022), f_{2021}(2023)\} = A,$$

where $\gcd(x, y)$ is the greatest common divisor of $x$ and $y$,
then

$$\frac{A}{5} = \, \underline{\hspace{1cm}}.$$

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Solution:

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We use the Euclidean Algorithm to solve this problem.

$$\gcd\{n + x^2, n + (x + 1)^2\} = \gcd\{n + x^2, n + x^2 + 2x + 1\} = \gcd\{n + x^2, 2x + 1\}.$$

Note that $2x + 1$ is odd, so multiplying $n + x^2$ by 4 will not change the greatest common divisor. Then we have

$$\gcd\{n + x^2, 2x + 1\} = \gcd\{4n + 4x^2, 2x + 1\} = \gcd\{4n + (2x + 1)^2 - 4x - 1, 2x + 1\} = \gcd\{4n - 4x - 1, 2x + 1\}$$ $$\gcd\{4n - 4x - 1, 2x + 1\} = \gcd\{4n + 1, 2x + 1\}.$$

Then the calculations are simple. $\gcd\{8005, 4005\} = 5$, $\gcd\{8045, 2025\} = 5$, and $\gcd\{8085, 2045\} = 5$, so the product is $5 \times 5 \times 5 = \boxed{125}$.

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Q29. (numbertheory)

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How many ways can $2^{2012}$ be expressed as the sum of four (not necessarily distinct) positive squares?

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Solution:

Q30. (geometry)

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A triangle $\triangle ABC$ is situated on the plane and a point $E$ is given on segment $AC$. Let $D$ be a point in the plane such that lines $AD$ and $BE$ are parallel.

Suppose that:

• $\angle EBC = 25^\circ$

• $\angle BCA = 32^\circ$

• $\angle CAB = 60^\circ$

Find the smallest possible value of $\angle DAB$ in degrees.

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Solution:

D could lie towards D' or D''.
If it lies towards D' then AD' and BE being parallel result in angle D'AB = 63.
If it lies towards D'' then AD'' and BE being parallel result in angle D''AB = 117.
So answer  = 63.