If n is a three digit number and last two digits of square of n are 54 (n 2 = … 54), then how many values of n are possible?

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Quant Warrior

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If n is a three digit number and last two digits of square of n are 54 (n^2 = … 54), then how many values of n are possible?

Please help me with this.
 
What ideas do you have? What have you tried? What techniques have you learned that might be relevant?

At the least, you might think about where the last two digits come from when you square a number. Or you might have a more powerful method (which might be overkill for this problem).
 
Dr.Peterson said:
What ideas do you have? What have you tried? What techniques have you learned that might be relevant?

At the least, you might think about where the last two digits come from when you square a number. Or you might have a more powerful method (which might be overkill for this problem).
Click to expand...

Following is the information I have about a perfect square ---

A perfect square cannot end with 2,3,7 or 8 and an odd number of zeroes. It can only end with 1,4,5,6 and 9 and an even number of zeroes.
The tens digit of a perfect square is always even except when the unit digit is 6.
(e1, e4, e5, o6, e9)

Every perfect square has only 22 different possibilities for last two digits which are deduced from squares 1 to 25.

As per above information, I can conclude that the tens digit of a perfect square has to be an even number if unit digit 4. So I think no such value of 'n' exits which will give last two digits as "54".

Above is my thinking. I am very raw . I have just started with basics of maths. So I feel a bit unsure about everything. (I am learning from the internet.)

Is my thinking correct ? Have I approached the question in a correct way or it was just a coincidence that the answer came. In my book, no solution has been given to that question. Only correct answers have been provides.

Please tell me all possible facts about this question.
 
If n^2 ends in 4, then n ends in a 2 or an 8.
Sometimes it is best to find the answer any way you can then think if there is a better way.
You can square 102, 122, 132,...992, 108,118,...998. That would involve squaring 180 numbers and will guarantee you that you find all the answers. Then armed with the answer think about doing this in a better way.
 
Maybe take the general 3 digit number, 100a + 10b + c and square it. Then you can equate the last two digits to 54. Possibly I gave away the problem
 
First, you seem to think that math is about knowing lots of facts. That is not correct. The number of numbers is infinite: it is impoosible to learn, let alone remember, all the pertinent facts about numbers. It is about knowing a few facts about numbers and some techniques and principles.

As I showed you in your last problem, you can break a large number down into a sum of smaller numbers and look for patterns that may be relevant.

[MATH]n = 100j + 10k + d[/MATH], where j, k, and d are non-negative integers less than ten and j is greater than zero.

The question is about the square of n. So what is that?

[MATH]n^2 = (100j + 10k + d)^2 = (100j + 10k)^2 + 2(100j + 10k)d + d^2 =\\ 10000j^2 + 2000jk + 100k^2+ 200jd + 20kd + d^2.[/MATH]Our original problem covered 900 numbers. We could use a computer to analyze them all. Or we can realize that we can greatly simplify the problem because we are interested only in the cases where

[MATH]20kd + d^2 = 100m + 50 + 4, [/MATH] where m is a non-negative integer..

Do you see what I have done? I started with the very GENERAL FACT that integers greater then 9 can be decomposed into a sum of products of one of ten small integers and powers of ten. And I then simplified the problem by identifying what is relevant. Now it is easy to figure out that 0^2 = 1, 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2 = 36, 7^2 = 49, 8^2 = 64, 9^2 = 81. Third grade arithmetic. No need to memorize something special.

So d = 2 or 8. What next?
 
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Evidently you are not approaching the problem with a general knowledge of number theory, but more or less "naively", so I'll take that perspective. In fact, I don't know your fact about the tens digit, so I won't assume that.

I'd start by just picturing how I'd calculate a square by hand:
Code: 
   xxx
*  xxx
------
  xxxx
 xxxx
xxxx
------
xxxx54

The only way to get the 4 in the units place is to multiply 2*2 or 8*8, so we have these possibilities (including a "carry"):
Code: 
                6
   xx2         xx8
*  xx2         xx8
------      ------
  xxx4        xxx4
 xxxx        xxxx
xxxx        xxxx
------      ------
xxxx54      xxxx54

Now I'd think about how we can get the 5.

This is more or less equivalent to what others have suggested, but without algebra.
 
Now we are sure about the values of "d''. It has to be either 2 or 8.

Let's substitute these values into 20kd and see when we can get 5 at ten's place.

20k×2 = 40k ( 40, 80, 120, 160, 200, .....

We always get a 0 at unit place no matter the value of k and also there is no 5 at tens place either.

So, 40k + 4

44,124, 164

As per above pattern, there is no 5 going to come at tens place.

And the same is the case for 20k ×8 = 160k.

If we put positive integers in place of k then 160k = 160, 320, 480, 640

160k + 4

Am I thinking in the right direction from where you left ?
 
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Dr.Peterson said:
Evidently you are not approaching the problem with a general knowledge of number theory, but more or less "naively", so I'll take that perspective. In fact, I don't know your fact about the tens digit, so I won't assume that.

I'd start by just picturing how I'd calculate a square by hand:
Code: 
   xxx
*  xxx
------
  xxxx
xxxx
xxxx
------
xxxx54

The only way to get the 4 in the units place is to multiply 2*2 or 8*8, so we have these possibilities (including a "carry"):
Code: 
                6
   xx2         xx8
*  xx2         xx8
------      ------
  xxx4        xxx4
xxxx        xxxx
xxxx        xxxx
------      ------
xxxx54      xxxx54

Now I'd think about how we can get the 5.

This is more or less equivalent to what others have suggested, but without algebra.
Click to expand...
Where and how can I get general knowledge of number theory ?
Regards
 
The best way is by taking a course in number theory! Lacking that, go to a university library and attempt to read a text book on number theory.
 
My own next step is to observe that the y has to be even, so the z has to be odd to make the 5. Can that happen?
Code: 
                6
   xx2         xx8
*  xx2         xx8
------      ------
  xxz4        xxz4
 xxxy        xxxy
xxxx        xxxx
------      ------
xxxx54      xxxx54

I'll let JeffM discuss the algebraic approach.
 
Quant Warrior said:
Now we are sure about the values of "d''. It has to be either 2 or 8.

Let's substitute these values into 20kd and see when we can get 5 at ten's place.

20k×2 = 40k ( 40, 80, 120, 160, 200, .....

We always get a 0 at unit place no matter the value of k and also there is no 5 at tens place either.

So, 40k + 4

44,124, 164

As per above pattern, there is no 5 going to come at tens place.

And the same is the case for 20k ×8 = 160k.

If we put positive integers in place of k then 160k = 160, 320, 480, 640

160k + 4

Am I thinking in the right direction from where you left ?
Click to expand...
Yes you are thinking sort of along the right lines, but you will notice that I said

[MATH]20kd + d^2 = 100m + 54.[/MATH]
Where in the world did that 100m come from?

Well, d might be 8, and k might be 9 so 20kd = 20 * 72 = 1440 > 1000. So I cannot presume that 20kd is under 1000. So actually, it would have been MUCH better had I said

[MATH]20kd + d^2 = 1000p + 100q + 50 + 4.[/MATH] Where p and q are non-negative integers. My previous statement was not corect.

Now if d = 2, we get

[MATH]20k(2) + 2^2 = 40k + 4 = 1000p + 100q + 50 + 4 \implies \\ 40k = 1000p + 100q + 50 \implies k = 25p + \dfrac{100q}{40} + \dfrac{50}{40} = 25p + \dfrac{10q + 5}{4}.[/MATH]which is not an integer if p and q are integers because 25p is an integer but (10q + 5)/4 is not an integer. BUT k is an integer by hypothesis. So d = 2 leads to a contradiction. We conclude d is not 2.

Now see what happens if d = 8.

As for number theory, it is possible to learn math on your own, but it is not easy to do so. You have not told us what your math training was or what you are curently studying. If it is something like foundations of mathematics that is not a place to start, nor is number theory.
 
Last edited:
JeffM said:
First, you seem to think that math is about knowing lots of facts. That is not correct. The number of numbers is infinite: it is impoosible to learn, let alone remember, all the pertinent facts about numbers. It is about knowing a few facts about numbers and some techniques and principles.

As I showed you in your last problem, you can break a large number down into a sum of smaller numbers and look for patterns that may be relevant.

[MATH]n = 100j + 10k + d[/MATH], where j, k, and d are non-negative integers less than ten and j is greater than zero.

The question is about the square of n. So what is that?

[MATH]n^2 = (100j + 10k + d)^2 = (100j + 10k)^2 + 2(100j + 10k)d + d^2 =\\ 10000j^2 + 2000jk + 100k^2+ 200jd + 20kd + d^2.[/MATH]Our original problem covered 900 numbers. We could use a computer to analyze them all. Or we can realize that we can greatly simplify the problem because we are interested only in the cases where

[MATH]20kd + d^2 = 100m + 50 + 4, [/MATH] where m is a non-negative integer..

Do you see what I have done? I started with the very GENERAL FACT that integers greater then 9 can be decomposed into a sum of products of one of ten small integers and powers of ten. And I then simplified the problem by identifying what is relevant. Now it is easy to figure out that 0^2 = 1, 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2 = 36, 7^2 = 49, 8^2 = 64, 9^2 = 81. Third grade arithmetic. No need to memorize something special.

So d = 2 or 8. What next?
Click to expand...
Now we are sure about the values of "d''. It has to be either 2 or 8.

Let's substitute these values into 20kd and see when we can get 5 at ten's place.

20k×2 = 40k ( 40, 80, 120, 160, 200, .....

We always get a 0 at unit place no matter the value of k and also there is no 5 at tens place either.

So, 40k + 4

44,124, 164

As per above pattern, there is no 5 going to come at tens place.

And the same is the case for 20k ×8 = 160k.

If we put positive integers in place of k then 160k = 160, 320, 480, 640

160k + 4

Am I thinking in the right direction from where you left ?

Last edited: 8 minutes ago
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Quant Warrior
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1 minute ago

Dr.Peterson said:
Evidently you are not approaching the problem with a general knowledge of number theory, but more or less "naively", so I'll take that perspective. In fact, I don't know your fact about the tens digit, so I won't assume that.

I'd start by just picturing how I'd calculate a square by hand:
Code:
xxx
* xxx
------
xxxx
xxxx
xxxx
------
xxxx54
The only way to get the 4 in the units place is to multiply 2*2 or 8*8, so we have these possibilities (including a "carry"):
Code:
6
xx2 xx8
* xx2 xx8
------ ------
xxx4 xxx4
xxxx xxxx
xxxx xxxx
------ ------
xxxx54 xxxx54
Now I'd think about how we can get the 5.

This is more or less equivalent to what others have suggested, but without algebra.
Click to expand...
Click to expand...
Where and how can I get general knowledge of number theory ?
Regards

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  • JeffM said:
    Yes you are thinking sort of along the right lines, but you will notice that I said

    [MATH]20kd + d^2 = 100m + 54.[/MATH]
    Where in the world did that 100m come from?

    Well, d might be 8, and k might be 9 so 20kd = 20 * 72 = 1440 > 1000. So I cannot presume that 20kd is under 1000. So actually, it would have been MUCH better had I said

    [MATH]20kd + d^2 = 1000p + 100q + 50 + 4.[/MATH] Where p and q are non-negative integers. My previous statement was not corect.

    Now if d = 2, we get

    [MATH]20k(2) + 2^2 = 40k + 4 = 1000p + 100q + 50 + 4 \implies \\ 40k = 1000p + 100q + 50 \implies k = 25p + \dfrac{100q}{40} + \dfrac{50}{40} = 25p + \dfrac{10q + 5}{4}.[/MATH]which is not an integer if p and q are integers because 25p is an integer but (10q + 5)/4 is not an integer. BUT k is an integer by hypothesis. So d = 2 leads to a contradiction. We conclude d is not 2.

    Now see what happens if d = 8.

    As for number theory, it is possible to learn math on your own, but it is not easy to do so. You have not told us what your math training was or what you are curently studying. If it is something like foundations of mathematics that is not a place to start, nor is number theory.
    Click to expand...
    As for d = 8

I got k = (100p + 100q +1)/ 16

I think this time k is not an integer either.


I think my level of maths is below an 8th grader. I am preparing for exams like GMAT which have aptitude test called Quantative Aptitude. Please guide me. Where should I start for a good foundtion of not just number theory but also of Quantitative maths?
 
Quant Warrior said:
Now we are sure about the values of "d''. It has to be either 2 or 8.

Let's substitute these values into 20kd and see when we can get 5 at ten's place.

20k×2 = 40k ( 40, 80, 120, 160, 200, .....

We always get a 0 at unit place no matter the value of k and also there is no 5 at tens place either.

So, 40k + 4

44,124, 164

As per above pattern, there is no 5 going to come at tens place.

And the same is the case for 20k ×8 = 160k.

If we put positive integers in place of k then 160k = 160, 320, 480, 640

160k + 4

Am I thinking in the right direction from where you left ?

Last edited: 8 minutes ago
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Quant Warrior
New member
1 minute ago


Where and how can I get general knowledge of number theory ?
Regards

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I got k = (100p + 100q +1)/ 16

I think this time k is not an integer either.


I think my level of maths is below an 8th grader. I am preparing for exams like GMAT which have aptitude test called Quantative Aptitude. Please guide me. Where should I start for a good foundtion of not just number theory but also of Quantitative maths?
Click to expand...
[MATH]20kd + d^2 = 1000p + 100q + 50 + 4 \text { and } d = 8 \implies \\ 160k + 64 = 1000p + 100q + 50 + 4 \implies 160k + 60 = 1000p + 100q + 50 \implies \\ 160k = 1000p + 100q - 10 \implies 16k = 100p + 10q - 1.[/MATH]But that implies that 16k is an odd number. And that implies that k is not an integer because the product of any integer and an even integer is an even integer. Again we have a contradiction. So d is neither 2 nor 8.

Going back to the original problem, what is the answer?

What you need to study is algebra. All the mathematics that you are ever likely to need (unless you intend on becoming a mathematician, a physicist, a software engineer, or a theoretical economist) requires algebra plus a tiny amount of geometry. Even if you studied it earlier, a review in algebra will be invaluable if you have the slightest doubt about your mastery of algebra. I have no idea what the GMAT covers, but calculus, probability theory, and statistics all require some facility with algebra. Calculus requires a lot of facility with algebra.
 
Quant Warrior said:
I think my level of maths is below an 8th grader. I am preparing for exams like GMAT which have aptitude test called Quantitative Aptitude. Please guide me. Where should I start for a good foundation of not just number theory but also of Quantitative maths?
Click to expand...
According to https://www.princetonreview.com/business/gmat-sections, the quantitative part "Tests general knowledge in arithmetic, basic algebra and basic geometry". You don't need number theory beyond what you need for fractions. Definitely work on algebra.

More details are listed here: https://www.gmatsyllabus.com/quantitative-section/gmat-math-topics. Again, master basic algebra!
 
JeffM said:
[MATH]20kd + d^2 = 1000p + 100q + 50 + 4 \text { and } d = 8 \implies \\ 160k + 64 = 1000p + 100q + 50 + 4 \implies 160k + 60 = 1000p + 100q + 50 \implies \\ 160k = 1000p + 100q - 10 \implies 16k = 100p + 10q - 1.[/MATH]But that implies that 16k is an odd number. And that implies that k is not an integer because the product of any integer and an even integer is an even integer. Again we have a contradiction. So d is neither 2 nor 8.

Going back to the original problem, what is the answer?

What you need to study is algebra. All the mathematics that you are ever likely to need (unless you intend on becoming a mathematician, a physicist, a software engineer, or a theoretical economist) requires algebra plus a tiny amount of geometry. Even if you studied it earlier, a review in algebra will be invaluable if you have the slightest doubt about your mastery of algebra. I have no idea what the GMAT covers, but calculus, probability theory, and statistics all require some facility with algebra. Calculus requires a lot of facility with algebra.
Click to expand...
d is neither 2 nor 8. But only sqaures of these two digits end in 4. But How is this fact related to our problem ? Please enlighten me further.
 
Quant Warrior said:
d is neither 2 nor 8. But only sqaures of these two digits end in 4. But How is this fact related to our problem ? Please enlighten me further.
Click to expand...
You agree that our three digit number n must end in 2 or 8 for the square of n to end in 4, right?

So if we divide 2* 2 * 10k = 40k by 10, we get the even number 4k, which can't be the odd number 5 if k is an integer. Any problem there?

And if we divide 2 * 8 * 10k + 60 = 160k + 60 by 10, we get the even number 16k + 6, which can't be the odd number 5 if k is an integer. Any problem there?

So neither d = 2 nor d = 8, which means that nothing works, and there is no such number n. The answer is zero.

Based on what Dr. Peterson said about the scope of the GMAT, I greatly doubt that a problem like this one will appear. If one does, it will not be critical to your score. This sort of touches on proofs by contradiction and existence proofs, which are what are deemed basic math.
 
Last edited:
[MATH][SUB][/SUB][/MATH]
JeffM said:
You agree that our three digit number n must end in 2 or 8 for the square of n to end in 4, right?

So if we divide 2* 2 * 10k = 40k by 10, we get the even number 4k, which can't be the odd number 5 if k is an integer. Any problem there?

And if we divide 2 * 8 * 10k + 60 = 160k + 60 by 10, we get the even number 16k + 6, which can't be the odd number 5 if k is an integer. Any problem there?

So neither d = 2 nor d = 8, which means that nothing works, and there is no such number n. The answer is zero.

Based on what Dr. Peterson said about the scope of the GMAT, I greatly doubt that a problem like this one will appear. If one does, it will not be critical to your score. This sort of touches on proofs by contradiction and existence proofs, which are what are deemed basic math.
[/QUO
But why did you divide by 10 ? Is it to check unit digit ( remainder) ?
Click to expand...
 
JeffM said:
You agree that our three digit number n must end in 2 or 8 for the square of n to end in 4, right?

So if we divide 2* 2 * 10k = 40k by 10, we get the even number 4k, which can't be the odd number 5 if k is an integer. Any problem there?

And if we divide 2 * 8 * 10k + 60 = 160k + 60 by 10, we get the even number 16k + 6, which can't be the odd number 5 if k is an integer. Any problem there?

So neither d = 2 nor d = 8, which means that nothing works, and there is no such number n. The answer is zero.

Based on what Dr. Peterson said about the scope of the GMAT, I greatly doubt that a problem like this one will appear. If one does, it will not be critical to your score. This sort of touches on proofs by contradiction and existence proofs, which are what are deemed basic math.
Click to expand...
Why did you divide by 10? (Is it to check unit place digit)
 
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