Since last 3 digits of $N^{2}$ are $N$ itself, $N^{2}-N=N(N-1)$ is divisible by $1000=2^{3}.5^{3}$ . But it is given that $N$ is even. Hence we conclude that $N$ is divisible by $8$ and $N-1$ is divisible by $125$ . Going through individual cases we get $N=376$ .

We need $N^2\equiv N\pmod{1000}$ , so $N(N-1)\equiv0\pmod{8}$ and $N(N-1)\equiv 0\pmod{125}$ . Only one of $N,N-1$ can be even, so that one must be divisible by $8$ . Thus $N\equiv 0,1\pmod{8}$ . Since $N$ is even, we must have $N\equiv 0\pmod{8}$ specifically.

Similarly $N\equiv 0,1\pmod{125}$ because only one of $N,N-1$ can be divisible by $5$ . If $N\equiv 0\pmod{125}$ , then $1000|N$ , so $N$ can't be a three-digit number. Thus $N\equiv 1\pmod{125}$ , and solving with $N\equiv 0,1\pmod{8}$ yields $N=376\pmod{1000}$ , so the value of $N$ must be $\boxed{376}$ .

N^2 = 1000a + N, (where a can be any positive integer)

therefore N^2-N = 1000a N(N-1)= 1000a => 1000 divides N(N-1)

We have to split the factors of 1000 (namely 2^3 and 5^3) among the 2 consecutive numbers. now we can't have powers of 2 on each number, because that would mean that the numbers will now both be even and therefore non-consecutive.

So one number is divisible by 2^x where x>=3

We cant place a power of five on the number which is divisible by 2^x where x>=3, because that would mean this number ends in 0. and that implies that its consecutive number either ends in 1 or 9 which can not be achieved by the remaining number as it can only end in 5 or 0.

Therefore one number is an ODD multiple of 125, and the other is divisible by divisible by 2^x where x>=3.

The question is asking for an N that is even, so N is not the odd multiple of 125 (since that is always odd). so we conclude that N-1 is the odd multiple of 125

so (N, N-1) (126, 125) 126 is not divisible by 8 (376, 325) 376 IS divisible by 8 (626, 625) 626 is not divisible by 8 (876, 875) 876 is not divisible by 8

N = 376

I have always known this special property: when you have an integer that ends with 76, you get back 76 as the last two digits of the product you obtain if you divided your number with another integer having 76 as its last 2 digits.

For example, say you have 2076 which ends in 76 and say, another number is 76 which obviously ends in 76. If you multiply 2076 with 76, you get 157776 which also ends in 76! This special property of 76 is which we shall use here.

Clearly, as the last digit of 76 is 6, which is even, we can say our solution is a three digit number ending in 76. This means the next work is trivial...to find the hundreds digit which one can even do using a calculator. However, since this is a mathematics site, we shall try to find x in "x76 " without using calculators.

Our number is x76, say, which means if we square it, we obtain (100x + 76)^2 = 10000x^2 + 15200x + 5776, simply using the expansion (a+b)^2 = a^2 + b^2 + 2ab.

Next, we observe that as we need only the last three digits, we may safely leave out 10000x^2 and the 5 of 5776. So we are left with 15200x + 776.

Here, we find two equations: 2x + 7 = x ... (1) or 2x + 7 = x + 10...(2). x will satisfy any one of these equations, but not both, simultaneously, as is obvious from number theory concepts.

From (1) we get x = -7 but x being a natural number will never take negative value.

From (2) we have x = 3.

So, our number is 376.

Let $N=XYZ$ for $X=\{1,2,...,9\}$ , $Y=\{0,1,2,...,9\}$ and $Z=\{0,2,4,6,8\}$ . The following conditions must be met:

$Z\equiv Z^2 \bmod{10} \Rightarrow Z=6$ , since for $Z=0$ it turns out that $X,Y=0$ which is problematic as $000$ not a 3-digit number.

$Y\equiv [(6Y+3)\bmod{10}+(6Y)\bmod{10}]\bmod{10} \Rightarrow Y=7$ . Note that since $(6Y+3)$ is odd, we must only check this constraint for the odd possible values of Y.

$X=[(6X+4)\bmod{10}+3+(6X)\bmod{10}]\bmod(10) \Rightarrow X=3$ .

It all becomes apparent once we understand what it means for $N^2\bmod{1000} \equiv N$ . Our final answer is $N=376$ .

Suppose we have such an $N$ . Let's label the digits of our number as $N = 100 \cdot a + 10 \cdot b + c = abc$ (I'd like to note that I don't mean the product of $a, b,$ and $c$ .

Note that the last digit of $N^2$ will be the last digit of $c^2$ . So we need for $c$ to be even so that squaring it does not change its last digit. What even numbers does this allow? Only $0$ and $6$ .

If $c =0$ , then in particular $10$ divides $N$ , so that $100$ divides $N^2$ . This means that the penultimate digit of $N^2$ is also a $0$ , so that $b = 0$ . But then $100$ divides $N$ , so that $1000$ divides $N^2$ . This means that $N = 000$ , which we discount. So we conclude that $c = 6$ .

Now we want to consider $b$ . Notice that the penultimate digit of $N^2$ is determined by the last two digits of $N$ . In particular, carrying out the multiplication, $\begin{array}{cccc} & & b & 6 \\ \times & & b & 6 \\ \hline = & & 3 + 6b & 6 \\ + & b^2 & 6b & 0 \\ \hline = & b^2 & 12b + 3 & 6 \end{array}$ where I abuse my notation just a bit (we don't actually care about the hundreds place yet, and there may be carry-over digits from $12b + 3$ in the tens place). But the important part is that the tens digit must be the same, so the last digit of $12b + 3$ must agree with $b$ . A quick check shows that the only possible number is $7$ , so we conclude that $b = 7$ .

All we have left is to find the hundreds digit $a$ . On the one hand, you could just check them - there are only ten possibilities. But let's see how to work it out. We carry out our multiplication again: $\begin{array}{cccc} & a & 7 & 6 \\ \times & a & 7 & 6 \\ \hline = & 6a + 4 & 5 & 6 \\ & 3 & 2 & 0 \\ & 6a & 0 & 0 \\ \hline = & 12a + 7 & 7 & 6 \end{array}$ where I again abuse notation and ignore all but the last three digits. So we need the last digit of $12a + 7$ to be the same as $a$ . Another quick computation shows that this means that $a = 3$ .

Thus we have the answer: $378$ .

Let the number be xyz.

By trial, there are 2 possibilities for z i.e 0 or 6. Let z=6. then for first stage multiplication,

xy6

xy6

=============

         (6y+3)     6

6y 0

+3 is the carried number as 6*6=36 hence 3 gets carried.

hence 6y+3+6y should have unit digit as y.

or 12y+3 should have unit digit as y.

again by trial (putting y as 0,1,2,....9 we get y should be 7.

now the multiplication becomes.

x76

x76

===============

 (6x+4)        5      6

 3             2      0

6x 0 0

+4 is the carried number as 6*7=42 so 4 gets carried.

now, 6x+4+3+6x should have unit place as x.

again by trial, we get x=3.

hence the number is 376.

now if z=0, and we do the same procedure, we will get the number as,

000.

which is not possible hence the answer is 376.

P.S---> I've used the basic way of multiplying i.e. something that we use when we are introduced to multiplying.

N can be expressed as 100a + 10b + c.

The units digit of N^2 must be c. Thus, the possible digits are 0 and 6. But if N will end with 0, then N^2 would end in 00. If N^2 ends in 00 then N will also end in 00. That being said, N^2 would then have to end in 0000 and N could not be 000. Thus, 6 is the only possible units digit.

Using trial and error for all b (1,2,3...9), only 76*76 will end up with a number with units digits 76.

Let x be the a number such that N^2 = x + 100a + 76 (100a + 76)^2 = x + 100a + 76 10000a^2 + 15200a + 5776 = x + 100a + 76 Only 15200x and 5776 will affect the value of the third digit. We must find an a such that the hundreds digit of (15200a + 5776) and (100a+76) will be the same. Thus, a = 0, 3 and since a cannot be 0, a = 3.

N = 376.

In order to find the three digit,Consider for digit at unit place i-e the digit should be an even and its square contains itself at unit place here 6 6=36 and 0 0=0 satisfy it. Then consider for digit at tens place here no two digit number end with 0 have its square end with same digits.Only 76 is two digit number end with 6 i-e 76 76=5776. Now consider for digit at hundred place,if we check orderly from 1 to 9 putting at hundred place with 76 then only 376 satisfy this condition i-e 376 376=141376. So,the only even three digit number is 376.

first select one digit number where,when you square it,the squared numbers,last digit is the number it's self.That wiil be 6.Then pick a two digit number by making 6,as the last digit numbers, where when you square it the last 2 digits are the same.that is 76.then go further by choosing a 3 digit by making 76,as the last 2 digit.were the last 3 digits make the number itself,that is 376.

First we can start off with the one's place value. In order for the one's digit of N be the same as the one's digit of $N^2$ , the one's digit can only be 4 possible numbers: $0, 1, 5,$ or $6$ . i.e. only 4 numbers $b$ can be expressed as $b^2 = 10a + b$ , where $a$ and $b$ are integers between $0$ and $9$ inclusive. $0^2 = 10 \times 0 + 0$

$1^2 = 10 \times 0 + 1$

$5^2 = 10 \times 2 + 5$

$6^2 = 10 \times 3 + 6$

However, we are given that $N$ is a positive number, and so the last digit cannot be $1$ nor $5$ ; it must be either $0$ or $6$ .

Moving into the ten's place value: If the last digit is a $0$ , then the last 2 digits can be written as $(10a + 0)^2 = (10a)^2 = 100a^2$ , meaning the last 2 digits of that number would both be 0s. Therefore, if the last digit of $N$ is $0$ , the second last digit must also be $0$ . If N ends in two 0s (divisible by 100), it can be written as $100(a)$ . $N^2$ would then be $(100a)^2 = 10000a^2$ , meaning the thousandth's digit would have to be 0. This is not possible, as 000 is not a three digit number.

Thus, we know the last digit of $N$ must be $6$ . A 2 digit number ending in $6$ can be written as $(10a + 6)$ , and so a 2 digit number ending in $6$ , squared can be written as $(10a + 6)^2 = 100a^2 + 120a + 36$ .

To find the ten's digit of $N^2$ , we look at the part $120a + 36$ . The $100a^2$ is not important as any integral value of $a$ would not affect the ten's digit.

In this case, we need $120a + 36$ to be equal to $100b + 10a + 6$ . Solving for a, we get $11a = 10b - 3$ . $10b - 3$ must end in a $7$ , therefore the ten's digit of $N$ , $11a$ must end in $7$ . Only the number $7$ satisfies this condition. Therefore the ten's digit is $7$ and we know N must be equal to $100c + 70 + 6.$

Again, for $N^2$ to have $N$ as the last 3 digits, $(100c + 76)^2 = 1000d + 100c + 76.$

Then we simplify:

$100c^2 + 151c + 57 = 10d$

$151c + 57 = 10(d-10c^2)$

$151c = 10(d - c^2) - 57$

$151c = 10(d - c^2 - 5) - 7$

As $d$ and $c$ are integers, the right hand side of the equation ends in 3. This means that the thousand's digit, $c$ , multiplied by $151$ must end in 3. Only the number 3 satisfies this condition, and thus it is the thousand's digit.

The number $N$ is thus $100(3) + 10(7) + 1(6)$

$300 + 70 + 6 = 376$

$376^2 = 141376$

since N is even and last three digits of N^2 is N, last digit of N must be 0 or 6. But if the last digit is 0, the penultimate digit will also be 0 and hence N must end with 2 zeroes. This narrows down the case to N being a multiple of 100 but none of those numbers satisfy the given condition. So N must end in 6. Now, consider P to be a 2 digit number ending in 6. So P^2 must have last 2 digits as P. This is satisfied only by 76 as 76^2 = 5776. Hence, the last 2 digits of N are 7 and 6. This gives 10 possible values of N out of which only 376^2 ends in 376. Hence N=376.

Let a,b,c be numbers from 0-9. Then, the square of a 3 digit number can be represented as (100a+10b+c)^2. Expanding the term, it is equal to 10,000a^2+100b^2+c^2+2000ab+200ac+20bc. Since the 3 digit number is even, c must be even. Now, we need to find c such that if we are going to square it the last digit is also c. The only possibility is 0 and 6, But 0 is not possible because, when the first digit of 3 digit number is zero, the second digit when we are going to square it is also zero. And if we let the tens digit of the 3 digit number be equal also to zero, then the square of the 3 digit will yield 4 zeros as the last digits, which is now impossible. Thus, 6 is the only possible number for c. Replacing c as 6, 10,000a^2+100b^2+36+2000ab+1200a+120b. Analyzing the new form, 36 and 120b are the only terms that would affect the second term of the square of the 3 digit. Thus, we need to find b, such that 120b+36 is equal to a term that have ending b and 6. But the only possible number for b from 0-9 is 7, thus b=7. substituting b as 7, then we have 10000a^2+4900+36+14000a+1200a+840 or simplifying further 10000a^2+15200a+5776. Now, 15200a+5776 are the only terms that could affect the hundred digit of the square of the 3 digit. Thus, we need to find a number for a from 0-9 that would yield 15200a+5776 to be a term whose ending is a, 7 and 6. But the only possible number is 3, hence a=3. Since, we let the square 3 digit number be represented as (100a+10b+c)^2, which now equal to (376)^2. The 3 digit number is 376.

This was a challenging one, not so much math as pattern recognition. I started out defining some variables

• N = abc
• $N^{2}$ = XYZabc

This doesn't actually do much, but it helps me visualize the problem, next I squared abc > abc * abc = (10,000) $a^{2}$ + (1,000)2ab + (100)( $b^{2}$ +2ac) (10)2bc + $c^{2}$

Since it's only the last digits that really matter, I started with those. $c^{2}$ has only one term to think about, making it an easy candidate. There are only a few numbers which when squared end in the same digit.

> c = {0,1, 5, 6} | $c^{2}$ = {0, 1, 25, 36}

N being an even number narrows it down

> c = {0 ,6}

Logically c cannot be 0, because if it were N^2 would end in "00", thus making b = 0, and by that same logic, making a = 0. Since the 3-digit number cannot be less than 100, this is absurd.

> c = 6

And c^2 is 36, meaning we have to carry the 3 over to the 10's place. Speaking of which let's examine that term.

> 2bc +3 = b (mod 10)

I'm doing modular division because I only care about the last digit

> 2(6)b + 3 = b (mod 10)

> 12b + 3 = b

Is equivalent to:

> 2b +3 = b (mod 10)

> b = -3 (mod 10)

Excuse my formatting but this is the way I think. The remainder can't be negative, so just add 10

> b = 7

And I have that answer. Now time to look at the 3rd digit, remember, a 8 carries over from the last term (7*12 + 3) = 87

> b^{2} + 2ac + 8 = a (mod 10)

> 7^{2} + 2(6)a + 8 = a (mod 10)

> 12a + 57 = a (mod 10)

> 2a + 7 = a (mod 10)

> a = -7 (mod 10)

> a = 3

There you go all 3 digits, abc = 376.

Only even number when squared gives same least significant digit is 6. Then, $(100 a_2 + 10 a_1 + 6 )^2 \equiv 100 (12 a_2 + a_1^2) + 120 a_1 + 36 \ (\mod 1000)$ We want this equal to $100 a_2 + 10 a_1 + 6 \ (\mod 10)$ . Equating and simplifying, we get, $1100 a_2 + 100 a_1^2 + 110 a_1 + 30 \equiv 0 \ (\mod 1000)$ This immediately gives, $a_1 = 7$ and $a_2 = 3$ .

Let A , B , C represent the hundreds, tens, and ones digits respectively of the number N .

Find C

From the problem description, N is equal to the last 3 digits of $N^2$ , which means that C is equal to the ones digit of $C^2$ . N is an even number, which in turn means C is even too. C = 6 is the only even number that satisfies this.

Find B

  AB6
  AB6
x__3_
    6

When performing multiplication, the tenths digit in this case is B = (6 B + 3) + 6 B , where the 3 is carried over from multiplying $6 \times 6$ .

B = 12 B + 3 (mod 10) .

This equation only holds true for B = 7

Find A

Now that we have B and C , we can write $N^2$ as $(100A + 76)^2$

= 10000 $A^2$ + $2 \times 100A \times 76$ + $76^2$ , The first term will have 3 zero's at the end, so we can safely ignore.

= 15200 A + 5776

Now we need to determine an A value such that

100 A + 76 = 15200 A + 5776 (mod 100)

A = 3 is the only value that satisfies this equation.

Thus N = 376

Using the quadratic formula (AB6)^2 = (100A+10B+6)^2, we are to find something that is ended with AB6, too or for easier sake, use calculator, try 3 digit number ended by 6, which if squared, will result a number with the last 3 digit is the origin number itself

$N$ is an even number, so the last digit of $N$ is even. Lets name it $x$ . The last digit of $x^{2}$ is equal to $x$ . So $x$ can be only $6$ or $0$ . It's not $0$ , because then $N^{2}$ would have last two digits: 00, so $N$ would end with 00, and then $N^{2}$ would end with 0000. $N$ is a 3-digit number, not zero. So $x=6$ . Now lets name the second digit $y$ . If last 3 digits of $N^{2}$ are equal to $N$ , then $y$ should be equal to the last digit of the sum of last digit of $2xy$ and the tenths digit of $x^{2}$ . So $y=2\times 6y+3$ ( $6^{2}=36$ ). After checking the only correct answer for $y$ is $7$ . Now time for the third digit. Lets name it $z$ . As before with $y$ , $z$ is equal to the last digit of the sum of last digit of $2zx$ , last digit of $x^{2}$ , and tenths digit of $2xy+3$ . So $z=2\times 6z+7^{2}+2\times 6\times 7+3$ . The only correct number for $z$ is $3$ . Now we get $x=6$ , $y=7$ and $z=3$ . After checking we get $376^{2}=141376$ .

I felt so good when I got it.. took about half an hour. At first I thought it was 625 and felt like smashing the keyboard when I read it had to be even.

Set $N=\overline{abc}$ . Because of last 3 digits of $N^2$ are N itself and N is even number, so c=0 or 6. But if c=0, we get b=0, so we get a=0. If c=6 we have , $N^2=(\bar{abc})^2=a.10^4+b.10^2+2.10^3ab+12.10^2.a+10^2.b+20b+36$ We note that " last 3 digits of $N^2$ " again, we have $2b+3= b$ or $2b+3=10+b$ Thus, $b=7$ . Now,we can try a=1,..9 to find solution for that problem. We get answer is $\boxed{376}$

I did this by trial and error. Note that the last digit must be $6$ ; since this is the only even number, when squared, returns the units digit as itself (other than 0, of course). I then tested $d6^2$ , where $d$ was a random digit between 1 and 9 to find another number which also satisfies the property. I found that 76 satisfies that property ( $76^2=5776$ ). From there, I tested again with digits for the hundreds unit; 3 matched the criterion ( $376^2=141376$ ). So $N=376$

Let the digits of the number be a , b and c. So, it is very clear that c = 6 as the number must be even and the last three digits of the square of the required number are the digits of the number itself. And as 6^2 = 36, c=6.--------------------------------------------- Let us look on the long multiplication.-------------------------------------------- -------------------------------------3--------------------------------------------------------------------------------------------a-b-6---------------------------------------------------------------------------------------x-a-b-6-------------------------------------------------------------------------------------_ _ _ _ _ _ _ _ _ _ _ ---------------------------------------------------------------------------------------------6------------------------------------------------------------------------------------ _ _ _ _ _ _ _ _ _ _ ---------------------------------------------- So, now b can only be 7 as only then will the condition satisfy which can easily be found using long multiplication.---------------------------------------- So, now,------------------------------------------------------------------------------- ------------------------------5--4-------------------------------------------------------------------------------------------4--3-------------------------------------------------------------------------------------------a--7--6------------------------------------------------------------------------------------x--a--7--6------------------------------------------------------------------------------------ _ _ _ _ _ _ _ _ _ _ ------------------------------------------------------------------------------------------5--6---------------------------------------------------------------------------------------3--2--x-------------------------------------------------------------------------------------------x--x---------------------------------------------------------------------------------- _ _ _ _ _ _ _ _ _ _ _ _-------------------------------------------------- Now, a must be 3 to get the condition satisfied which we can easily find using this long multiplication.------------------------------------------------ So, a=3 , b=7 and c=6-------------------------------------------------------------------So, the number is 376 whose square is 141376.

include<stdio.h>

void main () { long int i,n; for(i=100;i<=1000;i++) { n=i*i; if((n%1000)==i) { printf("\nRequired number=%d",i); } } getch(); } This returned the value of 376.

If you look on the even unit digits - 0,2,4,6 and 8, only 0 and 6 has the chance that if you square a number with 0 or 6 as their last digit, the resulting number has also 0 or 6 as the last digit. So, using a two-digit number with 0 as the last digit: 10^2=100, 20^2=400,......,90^2=8100. The pattern is the same, the last two digits are all 00. Then using 6: 6^2=36,16^2=256, 26^2=676, 36^2=1296, 46^2=2116, 56^2=3136, 66^2=4356, 76^2=5776, 86^2=7396, 96^2=9216. A pattern emerges but only 76^2 has a resulting last two digit equal to itself. So using three digit numbers with 76 as the last two digit: 176^2=30976, 276^2=76176, 376^2=141376,.....,976^2=952576. If we take all the squares from 176 to 976, we will find that only 376 has a resulting last three digit equal to itself. So the answer is 376.

Ok so first of all I am a 13 year old boy (It says 14 on my profile but I had to lie about my age. Hahaha) and I don't know much about math, so I need to use "creativity" or other methods to solve math problems so don't criticize me about posting a solution that has "no algebraic method involved."

So first I know by memory that 6 is the only digit that when squared, has the last digit as itself. So the last number of N must be 6. Starting at 16, I squared all the numbers 1-100 that have a 6 in the ones place to see if any of the numbers followed the criteria for the number N. I eventually found that 76 did. Using that method again, I found my answer, 376. :D Yay for trial and error.

376 x 376 = 141376