How many odd numbers between 10 and 1,000

The square of an odd number is an odd number:

10 < odd < 1,000
10 < odd^2 < 1,000
3.something < odd < 31.something (by taking the square root).

So, that odd number could be any odd number from 5 to 31, inclusive: 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, and 31.

14 numbers.

Answer: C.
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Hi All,

In these types of situations, you don't need to physically figure out every number in the set - you just have to figure out the smallest and the biggest, then you can 'count up' the number of terms in between (inclusive).

Here, we're asked for ODD numbers, between 10 and 1,000, that are SQUARES of INTEGERS. Since we're looking for ODD numbers, we're looking for the SQUARES of ODD numbers....

It helps to have certain perfect squares memorized and it helps to be able to do some basic multiplication by hand...

The first few perfect squares are....0, 1, 4, 9, 16, 25....

From this, we can determine the SMALLEST number that fits the above description: 25 (which is 5-squared). Now we have to figure out the BIGGEST number....

Here's where having strong arithmetic skills comes in handy (we can start with 'round' numbers to 'zero in' on the BIGGEST number)....

20^2 = 400 (which is way too small)
30^2 = 900 (which is getting close to 1,000)
31^2 = 961 (which is pretty close to 1,000)
33^2 = 1,089 (which is TOO BIG).

Now we know that 31-squared is the largest ODD number that fits the given description....Now we just have to count up the number of terms....

The ODD numbers are...
5-squared
7-squared
9-squared

11-squared
13-squared
15-squared
17-squared
19-squared

21-squared
23-squared
25-squared
27-squared
29-squared

31-squared

That's a total of 14 numbers.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

We know that only squares of odd numbers are odd
1^2=1, 3^2=9, 5^2=25........31^2=961,33^2=1089
we can see that squares of numbers between 5 and 31, inclusive, are the odd numbers lying between 10 and 1000
to find out how many numbers lie between 5 and 31, we may use sequences formula
nth term=first term + (no of terms - 1)*common difference
we have 31=5+(n-1)*2
Solving we get n=14
Hence Answer is choice (C)

Could you please advice on how to promptly figure out the upper and lower numbers for the set of odd number squares between 10 and 1000? Generally, how do you go about these types of questions?

Total Perfect Squares from 1 through 1000 = 1^2 to 31^2 i.e. 31 Perfect Squares (with 15 even and 16 odd perfect squares)

Odd Perfect Squares from 1 through 1000 = 31 - 15 = 16 perfect Squares

Odd Perfect Squares from 1 through 10 = 1^2 and 3^2 = 2 perfect Squares

Total Odd Perfect Squares from 10 though 1000 = 16 - 2 = 14

Answer: option C

You find the closest square you can think of and then go from there.

This is what I do when I want to find the first odd square after 10.
I know that 9 is the square of 3. (an odd square)
Next square of a number will be 4^2 which is 16 but this is an even square.
So 5^2 = 25 will be the odd square closest to but greater than 10.

Similarly, when I want the odd square closest to but less than 1000, I will think about 900, the square of 30 (It is very easy to see squares of numbers ending in 0 or 5). It is an even square less than 1000. The next number 31 will have an odd square.
On calculating, I find that 31^2 = 961.
The square of 32 is 1024 and the square of 33 will be even higher.

Hence I get the range 5 to 31.
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We have to find the squares at the extreme ends...
10----- square of 3 is 9.., so 4 is the lowest number in the series..
1000.... clearly it is above 30^2=900.... check for 31, it is 961 and then it goes above 1000..

So we are looking for ODDs from 4 to 31...
Total even and odd 31-4+1=28..
Odds =28/2=14
C
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Why don't we consider the negative integers here? For eg. 25 is the square of 5 and negative 5 also, so therefore in all there should be 14*2 that is 28 integers, right?

That is true but you have to count the number of perfect squares, not the number of square roots. Whether 25 is the square of one or two numbers, it doesn't matter. It is a perfect square and counted only once.
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I started by listing out the first few squares I knew:

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
5^2 = 25
6^2 = 36
7^2 = 49

At this point I realized that there was a pattern - ever other square was odd (because square of an odd is odd, and every other integer is odd).

To find upper boundary I knew 30^2 was 900 (based on 3^2 = 9 and adding two zeros for each of the 10's from each 30).

Instead of multiplying to test, I knew that 32^2 = (2^5)^2 = 2^10 = 1024, and therefore was too big. Because the difference between 1000 and 1024 is greater than 30, I assumed that the next integer down 31, was less than 1,000.

Then I counted odds between 5-31, to get 14.

or use ((31 - 5)/2) + 1 = 14

Answer C

The smallest odd perfect square between 10 and 1000 is 5^2 = 25.

The largest odd perfect square between 10 and 1000 is 31^2 = 961.

The number of odd integers from 5 to 31 inclusive is (31 - 5)/2 + 1 = 14.

Answer: C
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3*3=9<10 --> do not count 3
4*4 = 6 > 10
31*31=961<1000
32*32=1024 --> do not count 32
Then we should count odd numbers from 4 to 31, it contains:
5 7 9 11 13 15 17 19 21 23 25 27 29 31 (total 14 numbers)

This problem can be done using 2 basic things.

10----1000
we know lowest odd integer square = 5
highest = 32*32 = 1024 so = 31

odd integers:
31 = 5 + (n-1)2
n = 14

how my noob self solved this...

first figure the lowest vaible solution, i.e. 5^2... then figure the last possible number idk why i started from 23^2 here but i evenutally got to 31^2. from there i did ((31-5)/2) +1 . plus 1 because both inclusive answer 14... i.e. C... i have a long way to go in trying to test my min-max cases could have saved like 30 seconds

Hi Bunuel,

This is a GMAT Official Mock - 4 Question. Can you please change the source from "Others" to "GMAT Prep"? The level is 650-700 only.

Thanks.
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Thank you! Added the tag.
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