JEE Mathematics 2023 Question with Solution

Given: A triangle is formed by the coordinate axes and the line $3x + 4y = 60$.

Find: The number of points $P(a,b)$ lying strictly inside the triangle such that $a$ is an integer and $b$ is a multiple of $a$.

The intercepts of the line are $(20,0)$ and $(0,15)$ so the interior points satisfy $x>0, y>0$ and $$ 3x+4y<60

Since $$b$$ is a multiple of $$a$$, let $$b=ka$$ where $$k$$ is a positive integer.

Now count valid points for each integer value of $x=a$ using the bounds shown in the solution:

If $x=1$, then $$ y<\frac{60-3}{4}=\frac{57}{4}=14.25

So the points are $$(1,1),(1,2),\dots,(1,14)$$ giving **$$14$$** points.

If $x=2$, then $$ y<\frac{60-6}{4}=\frac{27}{2}=13.5

Multiples of $$2$$ below this are $$(2,2),(2,4),\dots,(2,12)$$ giving **$$6$$** points.

If $x=3$, then $$ y<\frac{60-9}{4}=\frac{51}{4}=12.75

The points are $$(3,3),(3,6),(3,9),(3,12)$$ giving **$$4$$** points.

If $x=4$, then $$ y<\frac{60-12}{4}=12

Strictly inside means $$y<12$$, so the points are $$(4,4),(4,8)$$ giving **$$2$$** points.

If $x=5$, then $$ y<\frac{60-15}{4}=\frac{45}{4}=11.25

The points are $$(5,5),(5,10)$$ giving **$$2$$** points.

If $x=6$, then $$ y<\frac{60-18}{4}=\frac{21}{2}=10.5

The point is $$(6,6)$$ giving **$$1$$** point.

If $x=7$, then $$ y<\frac{60-21}{4}=\frac{39}{4}=9.75

The point is $$(7,7)$$ giving **$$1$$** point.

If $x=8$, then $$ y<\frac{60-24}{4}=9

Strictly inside gives only $$(8,8)$$, so there is **$$1$$** point.

If $x=9$, then $$ y<\frac{60-27}{4}=\frac{33}{4}=8.25

No positive multiple of $$9$$ is less than $$8.25$$, so there is no point.

Hence total number of points is

$$14+6+4+2+2+1+1+1=31$$

Therefore, the required number of points is $31$.

Given: $3x+4y=60$ bounds the triangle in the first quadrant.

Find: Count all points $P(a,b)$ with integer $a$ and with $b$ a multiple of $a$, lying strictly inside the triangle.

For a fixed integer $a=x$, the largest allowed value of $y$ is obtained from

$$y<\frac{60-3x}{4}$$

Now $b$ must be one of $x,2x,3x,\dots$ below that upper bound.

Thus for each integer $x$, the number of choices equals the number of positive multiples of $x$ that are strictly less than $\frac{60-3x}{4}$. Evaluating for $x=1$ to $9$ gives counts:

• $x=1 \Rightarrow 14$
• $x=2 \Rightarrow 6$
• $x=3 \Rightarrow 4$
• $x=4 \Rightarrow 2$
• $x=5 \Rightarrow 2$
• $x=6 \Rightarrow 1$
• $x=7 \Rightarrow 1$
• $x=8 \Rightarrow 1$
• $x=9 \Rightarrow 0$

Adding these,

$$31$$

So the correct answer is $31$.