Tens Digit 7

Let $N = 100a + b$ , where $a$ is the hundreds digit. Note that $(100a+b)^2 \equiv b^2 \pmod{100}$ , so the tens term of $N^2$ is independent of $a$ . To minimise $N$ , we let $a=1$ .

Consider the units digits of $b^2$ . Since $b^2 \equiv 0, 1 \textrm{ or } 4 \pmod{5}$ , then the units digit of $b^2$ can be $0, 1, 4, 5, 6 \textrm{ or } 9$ . Also, because $b^2 \equiv 0 \textrm{ or } 1 \pmod{4}$ and the tens digit of $b^2$ is $7$ , then the units digit of $b^2$ can be $2, 3, 6 \textrm{ or } 7$ (since only $72, 73, 76, 77 \equiv 0 \textrm{ or } 1 \pmod{4}$ ). Thus, the units digit of $b^2$ is $6$ . From this, we can work backwards to know that $b \equiv 1 \textrm{ or } 4 \pmod{5}$ and $b \equiv 0 \pmod{2}$ . Thus, $b \equiv 4 \textrm{ or } 6 \pmod{10}$ .

Trying out $b = 4, 6, 14, 16 \textrm{ and } 24$ , we find that $b=24$ is the only value that will make $b^2 \equiv 76 \pmod{100}$ . So the smallest $N$ is $124$ .

Let $N=\overline{abc}$ be a 3 digit number.Then $N^2\equiv(100a+10b+c)^2\equiv(10b+c)^2\equiv c^2\pmod{20}$ .Since it have $7$ ,an odd number, as it's tens digit,then $c^2\equiv r\pmod{20}$ ,where $10\leq r \leq19$ .We obtain $c\in\{4,6\}$ .Using modulo 100, $N^2\equiv20bc+c^2\pmod{100}$ .Putting $c=4$ ,we have $80b+16\pmod{100}$ .Then $80b$ have $6$ as its tens digit,which the smallest $b$ that satisfy this is 2.So for case $c=4$ ,the smallest is 124(since we can choose $a$ randomly).The latter case have 126 as the smallest number.Hence 124 is the smallest possible value of $N$

Let $N=100a+10b+c$ where $a,b,c$ are integers and $1\leq a \leq 9, 0\leq b \leq 9, 0\leq c \leq 9$ . Then $N^2=10000a^2+100b^2+c^2+2(1000ab+10bc+100ac)$ . Now let's focus on the tens digit of $N^2$ . The terms that determine the tens digit of $N^2$ are $20bc$ and $c^2$ . Since we want the tens digit of $N^2$ to be odd and the tens digit of $20bc$ is even, the tens digit of $c^2$ must be odd. This means that the only possible values of $c$ are $4$ and $6$ .

Case $1$ : Let $c=6$ . We have $20bc+c^2=120b+36$ The smallest such $b$ that produces $7$ in the tens place is $2$ . Hence $N=126$ .

Case $2$ : Let $c=4$ . We have $20bc+c^2=80b+16$ The smallest such $b$ that produces $7$ in the tens place is $2$ . Hence $N=124$ .

Since $124 < 126$ the smallest $N$ such that the tens digit of $N^2$ is $7$ is 124.

Let $N = \overline{abc}$ . Let $y = \overline{bc}$ . Then $N^2 = (100a+y)^2 = 10000a^2 + 200ay + y^2$ . This shows that the tens digit of $N$ is independent of $a$ , so to get the smallest $N$ , $a$ must be $1$ .

We're now reduced to finding the smallest non-negative integer $Y$ such that the tens digit of $Y^2$ is $7$ . Note that $8^2 < 70 < 79 < 9^2$ , $13^2 < 170 < 179 < 14^2$ , $16^2 < 270 < 279 <17^2$ , $19^2 < 370 < 379 < 20^2$ , $21^2 < 470 < 479 < 22^2$ . Also note that $23^2 = 529$ and $24^2 = 576$ , so the smallest $Y$ is $24$ .

Hence we have the smallest $N$ as $124$ .

$$N^2=(100x+10y+z)^2\equiv20yz+z^2\pmod{100}$$

So, the value of x does not matter and $$x_{min}=1$$

$$\text{Note that, the tens digit of } N^2 \text{ will be odd} \iff \text{ the tens digit of }z^2\text{ is odd }$$

So, z must be 4 or 6

If $$z=4, N^2\equiv (8y+1)10+6 \pmod{100}\implies 8y+1\equiv7\pmod{10}\implies y=2,7\text{ as } 0\le y\le9$$

If $$z=6, N^2\equiv (12y+3)10+6 \pmod{100}\implies 12y+3\equiv7\pmod{10}\implies y=2,7 \text{ as } 0\le y\le9$$

So, $$N_{min}=100\cdot1+10\cdot2+4$$

Notice that the tens digit of the square of a number $\overline{abc}$ is the same as that of the number $\overline{bc}$ ,

Using this principle, we can simplify this problem to finding a two-digit number whose square has a 7 as its tens digit.

since $8^2=64, 9^2=81$

$13^2=169, 14^2=196$

$16^2=256, 17^2=289$

$19^2=361, 20^2=400$

$21^2=441, 22^2=484$

this leads us to $24^2=576$ which has a 7 as its tens digit

taking the smallest possible hundreds digit 1, 124 is confirmed to be the final answer.

Let $N=100a+10b+c$ . Then, $$N^2=10000a^2+2000ab+200ac+100b^2+20bc+c^2.$$ We want to work with the expression $20bc+c^2$ since this is what determines the tens digit of $N^2$ . To minimize $N$ , let $a=1$ . Now we can proceed by casework.

Case 1: $b=1$ We need to find $c$ such that $c^2+20c$ has a tens digit of $7$ . We can empirically see that no $c\in\{1,2,\ldots,9\}$ satisfies this, so $b\neq 1.$

Case 2: $b=2$ We need to find $c$ such that $c^2+40c$ . The minimal $c$ that satisfies this is given by $c=4$ , so that $c^2+40c=176$ . Thus, $a=1, b=2, c=4$ , so $\boxed{N=124}$ .

We can show that the hundreds digit of $N$ [its leftmost digit] can not influence the tens digit of $N^2$ in any way.

Let $N=100a+10b+c$ for some integers $a, b$ and $c$ [ $1\leq a\leq 9, 0\leq b\leq 9$ and $0\leq c\leq 9$ ].

So, $N^2=(100a+10b+c)^2$ $=10000a^2+100b^2+c^2+2000ab+20bc+200ca$ .

We can see that the terms that have $a$ in them have at least $3$ zeros on their right end. So they won't contribute to the tens digit in any way. That means whatever $a$ is, it doesn't affect the tens digit. Since we're looking for the smallest value of $N$ , we want the lowest possible value of $a$ which is $1$ .

Again we see that the only part that could affect the tens digit of $N^2$ is $c^2+20bc$ . So, $c^2+20bc$ has to have $7$ as its tens digit for $N^2$ to have $7$ as its tens digit. We can immediately see that $c\neq 0$ .

Another possibility that we can rule out is $b=0$ . Because then, the tens digit will completely depend on $c^2$ . And there is no $c$ [ $1\leq c\leq 9$ ]such that $c^2$ has $7$ as its tens digit.

Whatever, $b$ or $c$ may be $20bc$ would always have an even number at its tens digit and zero as its units digit. This is because $20bc$ is always a multiple of $20$ and can be written as $2k\times 10$ where $k$ is an integer.

If $c$ was $1, 2$ or $3$ , $c^2$ wouldn't affect the tens digit because then $c^2$ would be less than $10$ . Then the tens digit will entirely depend on $20bc$ . We've seen in the previous paragraph that $20bc$ will always have an even number as its tens digit. So, in this case [ $c<4$ ], $N^2$ won't have $7$ as its tens digit.

So, $c\geq 4$ .

Another thing that we understand from here is that the tens digit of $c^2$ has to be odd since the tens digit of $20bc$ is even. As we want $7$ , an odd number, as the tens digit of $N^2$ , we need a number with odd-tens-digit to add with the even tens digit of $20bc$ to get it.

This narrows $c$ down to $4$ and $6$ [ $4^2$ has $1$ as its tens digit and $6^2$ has $3$ as its tens digit. No other $c$ where $4\leq c\leq 9$ has an odd tens digit.]

Now we're going to test these two values of $c$ . If $c=4$ , we need the the tens digit of $20bc=80b$ to be $6$ so that we can add it with the tens digit of $c^2=4^2$ [which is $1$ ] to get $7$ . This gives us two values of $b$ : $2$ and $7$ .

And if $c=6$ , we need the tens digit of $20bc=120b$ to be $4$ so that we can add it with the tens digit of $c^2=6^2$ [which is $3$ ] to get $7$ . This again gives us two values of $b$ : $2$ and $7$ .

So, for smallest $N$ , $a=1, b=2$ and $c=4$ .

And $N=124$ .

The question says that the last 2 digits must be
70, 71, 72, 73, 74, 75, 76, 77, 78, 79.
(But using the properties of squares) The last digit can't be 2, 3, 7 and 8.

So, 72, 73, 77 and 78 are REJECTED.

Unit digit of any number --------- Last 2 digits of square

1 --------- (even)1
2 --------- (even)4
3 --------- (even)9
4 --------- (odd)6
5 --------- 25
6 --------- (odd)6
7 --------- (even)9
8 --------- (even)4
9 --------- (even)1
0 --------- 00

This table gives the information of last 2 digits of square of ANY number. So, using this table we REJECT the last 2 digits of the square which are 70, 71, 73, 74, 75, 79 (as 7 is odd).

Therefore, the last 2 digits of $N^2$ should be 76.

Observation-1:We observe that the last 2 digits of $\overline{ab}$ [2 digit number] and 1 $\overline{ab}$ [3 digit number] are equal.

So, if we find the smallest 2 digit number having last 2 digits as 76, we can easily find the 3 digit number $N$ .

Observation-2: we find that the 2 digit number is 24.
Hence, the number $N$ is 124 [ Observation-1 ]

We only care about the tens digit of $N^2$ , since $N$ needs to be as small as possible, the hundreds digit of N must be $1$ . $N$ can be expressed as $N = \overline{1ab}$ . Then we have:

$N^2 = (100+10a+b)^2$ $= 10000+100a^2+b^2+2000a+20ab+200b$

Let $P = b^2 + 20ab$ . The tens digit of $P$ is also $7$ , so value of P must satisfy one of these conditions:

$(1) 70 \le P < 80 \\ (2) 170 \le P < 180 \\ (3) 270 \le P < 280 \\ etc.$

In the first case, with any value of $b$ from 0 to 9, it doesn't exit integer value of $a$ . In the second case, if $b = 4$ , then $a = 2$ . Hence, the answer is $124$ .

An attractive argument, but with a serious flaw: Minimizing Z is not guaranteed to minimize the number. Possibly feature as incorrect.

If you square a 3-digit number, the last two digits are always the same as the square of the 2-digit number itself.

Given an example : 12 x 12 = 144 and 112 x 112 = 12544

Back to the question, a 2-digit number which has 7 as its tens digit is 24, as 24 x 24 = 576.

So the smallest 3-digit number which has the same last two digits is 124 as 124 x 124 = 15376

the last two digits of N ^ 2 the same as the last two digits of the square of the last two digits N. so digit ratusannya not influence example: 214 ^ 2 = 45796 14 ^ 2 = 196 the same last two digits equal 96 so, just look for the square two digit 7 digit puluhannya ketemunya 24 ^ 2 = 576 124 ^ 2 = 15376 N = 124

Consider the quadratic residues modulo $100$ . The only instance when $7$ appears in the tens digit is when $x^2 \equiv 76 \pmod{100}$ . Hence, this tells us that the last digit of $N$ must be either $4$ or $6$ . Since $(x+100)^2 \equiv x^2 \pmod{100}$ , the hundreds digit of $N$ is $1$ , and we only need to check the 2-digit numbers.

For the 2-digit numbers, checking $4$ , $6$ , $14$ , $16$ , $24$ , we see that $24^2 = 576$ is the smallest integer that satisfies the condition. Thus, $N=124$ .

Just try from 100 onward and output the first valid solution. Note that the first digit of $N$ does not affect the tens digit of $N^2$ so we just need to square the last two digits when checking.