How many square numbers between \[1\] and \[1000\] are divisible by \[3\]?
Answer
657.3k+ views
Hint:
The Squares of numbers \[1\] to \[31\] lie between \[1\] to\[1000\]. Out of these numbers \[1\] to \[31\]only \[10\] numbers are divisible by \[3\]. Thus, their squares will also be divisible by \[3\].
Complete step by step solution:
In \[1\] to\[1000\], there are total\[333\] numbers those can be divided by \[3\]
But here we need to consider the squares of those multiples.
We know that only squares of multiples of \[3\] lying between \[1\] to\[31\] will fall in between \[1\] to \[1000\]. Now out of \[1\] to\[31\], there are only \[10\] numbers whose squares fall in between \[1\] to\[1000\] but they are multiples of \[3\]as well.
Basically, the square of \[31\] will not be a multiple of \[3\].
We have assumed, here, that a number is ‘divisible by \[3\]’ means ‘with a \[0\] remainder’. If that isn’t the case, then all \[31\] numbers would qualify and your count would be \[31\].
So square numbers divisible by \[3\] from \[1\] to \[1000\] are basically the square of the numbers those are divisible by 3 in between \[1\] to \[31\], we get:
\[9,{\text{ }}36,{\text{ }}81,{\text{ }}144,{\text{ }}225,{\text{ }}324,{\text{ }}441,{\text{ }}576,{\text{ }}729,{\text{ }}and{\text{ }}900\],
i.e. the numbers that are divisible by 3 and there are \[10\] such numbers.
Note:
Check the Numbers whose squares fall in between \[1\]to\[1000\]. These are from \[1\] to\[31\]. Identify Out of these which are multiple of \[3\] as well. If the square is divisible by \[3\], then the square root must also be divisible by \[3\]. The square root of \[1000\] is\[31.6\], so all you should have to look at are the numbers between \[1\] and \[31\] that are divisible by \[3\]. Count the ones divisible by \[3\] and there should be about \[10\] of them. We have assumed, here, that ‘divisible by\[3\]’ means ‘with a \[0\] remainder’.
The Squares of numbers \[1\] to \[31\] lie between \[1\] to\[1000\]. Out of these numbers \[1\] to \[31\]only \[10\] numbers are divisible by \[3\]. Thus, their squares will also be divisible by \[3\].
Complete step by step solution:
In \[1\] to\[1000\], there are total\[333\] numbers those can be divided by \[3\]
But here we need to consider the squares of those multiples.
We know that only squares of multiples of \[3\] lying between \[1\] to\[31\] will fall in between \[1\] to \[1000\]. Now out of \[1\] to\[31\], there are only \[10\] numbers whose squares fall in between \[1\] to\[1000\] but they are multiples of \[3\]as well.
Basically, the square of \[31\] will not be a multiple of \[3\].
We have assumed, here, that a number is ‘divisible by \[3\]’ means ‘with a \[0\] remainder’. If that isn’t the case, then all \[31\] numbers would qualify and your count would be \[31\].
So square numbers divisible by \[3\] from \[1\] to \[1000\] are basically the square of the numbers those are divisible by 3 in between \[1\] to \[31\], we get:
\[9,{\text{ }}36,{\text{ }}81,{\text{ }}144,{\text{ }}225,{\text{ }}324,{\text{ }}441,{\text{ }}576,{\text{ }}729,{\text{ }}and{\text{ }}900\],
i.e. the numbers that are divisible by 3 and there are \[10\] such numbers.
Note:
Check the Numbers whose squares fall in between \[1\]to\[1000\]. These are from \[1\] to\[31\]. Identify Out of these which are multiple of \[3\] as well. If the square is divisible by \[3\], then the square root must also be divisible by \[3\]. The square root of \[1000\] is\[31.6\], so all you should have to look at are the numbers between \[1\] and \[31\] that are divisible by \[3\]. Count the ones divisible by \[3\] and there should be about \[10\] of them. We have assumed, here, that ‘divisible by\[3\]’ means ‘with a \[0\] remainder’.
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