Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and $10$ less sharpeners. If the cost of one sharpener is Rs. $2$ more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

We need to set up equations based on the spending conditions and find the minimum number of pencils.

Let us define variables clearly:

Cost of one pencil = Rs. $x$

Cost of one sharpener = Rs. $(x + 2)$ [given that sharpener costs Rs. 2 more]

Aron's purchases:

Number of pencils = $a$

Number of sharpeners = $b$

Total spending = $ax + b(x + 2)$

Aditya's purchases:

Number of pencils = $2a$ [twice as many as Aron]

Number of sharpeners = $(b - 10)$ [10 less than Aron]

Total spending = $2ax + (b - 10)(x + 2)$

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Since both spent the same amount:

$ax + b(x + 2) = 2ax + (b - 10)(x + 2)$

Expanding the right side:

$ax + bx + 2b = 2ax + bx + 2b - 10x - 20$

Key insight: Notice that $bx + 2b$ appears on both sides, so they cancel out:

$ax = 2ax - 10x - 20$

$-ax = -10x - 20$

$ax = 10x + 20$

Dividing both sides by $x$ (assuming $x > 0$):

$a = 10 + \tfrac{20}{x}$

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For $a$ to be minimum, we need $\tfrac{20}{x}$ to be minimum.

Important constraint: Since $a$ must be a positive integer, $\tfrac{20}{x}$ must also be an integer. This means $x$ must be a divisor of 20.

The divisors of 20 are: 1, 2, 4, 5, 10, 20

To minimize $\tfrac{20}{x}$, we need to maximize $x$. The largest divisor is $x = 20$.

When $x = 20$:

$a = 10 + \tfrac{20}{20} = 10 + 1 = 11$

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Total pencils bought by both:

Aron: 11 pencils

Aditya: 22 pencils

Total: $11 + 22 = 33$ pencils

Why this is minimum: Any smaller value of $x$ would give a larger value of $a$, and any value of $x$ that doesn't divide 20 would make $a$ non-integer, which is impossible since you can't buy a fraction of a pencil.

Therefore, the minimum possible number of pencils bought by Aron and Aditya together is 33.