Please help me in the problem

Q: Determine the smallest counting number having 36 factors with 6 consecutive factors.

#NumberTheory

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

The answer is 1800.

Hint: By pigeonhole principle, show that the number you are looking for (let's call it $X$ ) is divisible by 3, 4 and 5.

Hint 2: Show that $X$ must be divisible by $2\times 4= 8$ .

Hint 3: Show that $X$ is divisible by $2^a \times 3^b \times 5^c$ for integers $a\geq 3, b\geq1, c\geq1$ .

Hint 4: Apply Number of factors formula. Express 36 as the product of (not necessarily distinct) positive integers greater than 1.

Hint 5: Show that $X$ only have 3 distinct prime factors. You will end up with 4 positive integers that satisfy the given constraints.

Pi Han Goh - 3 years ago

Thank you sir.

s p - 3 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$