What Number?

This is a question I am unable to solve. Please help me find the answer.

Find the least integer greater than 0 whose digit comprises of only 0s and 1s, and which is perfectly divisible by 225.

#NumberTheory

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Comments

First off a number is divisible by 225 if and only if it is divisible by both 25 and 9.

Now, note that a number is divisible by 25 if and only if the last two digits (last from the left) are divisible by 25 (for example 455739250 is divisible by 25 as 50 is divissible 25 and 5028564 is not as 64 is not divisible by 25).

So, the last two digits of the required number must be 0.

Now, a number is divisible by 9 if and only if the sum of the digits is divisible by 9.

Hence, there must be nine 1 s in the required number.

And so, the least number with the desired properties is:

$\large 11,111,111,100$

A Former Brilliant Member - 4 years, 11 months ago

Thank you very much for the simple explanation.

Nilabha Saha - 4 years, 11 months ago

You're welcome.

A Former Brilliant Member - 4 years, 11 months ago

Ans is 11,111,111,100

Hemanth K - 4 years, 11 months ago

Are you Hemath K of N.P.S. YPR. Please forgive me if it is not you.

Nilabha Saha - 4 years, 11 months ago

Genius @Bhat

Ayanlaja Adebola - 4 years, 11 months 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}`	$\boxed{123}$