Math, asked by nehasinghsuryawanshi, 1 month ago

find a 2 digit number and the number formed by interchanging its digits such that their squares are 2 number with the same digits but in reverse order​

Answers

Answered by selfiequeen001
0

Answer:

The figure shows the favourite sports of Grade 7 and Grade 8 students. Analyse and answer the questions.

A. Which grade likes table tennis the most? B. Which sport do the most Grade 8 students like?

C. Which grade has more interest in cricket?

D. how many students like football in total?

Step-by-step explanation:

Hy friend, thanx for asking the question...

Here's ur ans....

9025

5×5×19×19

We hv a pair of 5 and a pair of 19

Take one 5 and one 19 from it,

That is,

5×19

95

\sqrt{9025 } = \: 95

9025

=95

Hope it helped u...

Answered by roydhiraj1949
0

Let’s assume the number is (10x+y) when x and y are positive integers.

so its square is 100x^2+10(2xy)+y^2.——————[1]

And the reverse of the first number is (10y+x).

So, its square is 100y^2+10(2yx)+x^2.——————[2]

So,while reversing x^2 and y^2 will change their places and it will be all fine.

BUT,it only happens when they are limited in their digit’s places and do not make a carry over when added.

For example,14^2=100+80+16

and 41^2=1600+80+1.

Here the problem is that when 16 and 1 will interchange their places,they will cause a carry over of number and it will ruin the required answer.

Because,80+16=96 and 80+1=81

That’s the problem.

So,to make this equation applicable ,

x^2<10,y^2<10<and also 10(2xy)<100 or,(xy)<5

Now,the possible answers are 12[x=1<10 & y=2<10] , 13[x=1<10 & y=3<10] , 11 (x=y=1<10) and 22 (x=y=2 <10).

So the answers are

12–21,13–31,11–11,22–22

Hope this helps.

This is not best way but programmatically you can do it this way:

function findNumber() {

var revNumber,

square,

revSquare,

revSquareRoot,

i = 10;

for (; i <= 100; i++) {

revNumber = Number(i.toString().split('').reverse().join(''));

square = i * i;

revSquare = Number(square.toString().split('').reverse().join(''));

revSquareRoot = Math.sqrt(revSquare);

if (revNumber === revSquareRoot) {

console.log("Number: " + i);

}

}

}

True story:

About 50 years ago, when I was in school, a non-mathematical guy I knew was doing some calculations for a business statistics class that he was taking. This involved the tedious process of squaring a lot of numbers by hand. At one point he asked me “Walter, what is 31 squared?”

I saw the opportunity for a good practical joke, and pounced. I asked him if he knew the answer to 13 squared, and he said “Yes, 169.” I told him that 31 is the reverse of 13, so 31 squared would be the reverse of 169. He asked if that was really correct, and I said sure, work it out by hand and you will see that it is correct. He worked it out and verified that the answer was 961. He then asked if that process would always work and I said “Sure! 11 squared is 121, but that is a trivial example. 12 squared is 144 and 21 squared is 441. As we have already seen, 13 squared is 169 and 31 squared is 961. And so on.”

He continued, apparently happy that he had found a short-cut.

I saw him a couple of days later and he complained that he had wasted a lot of time because the “short-cut” I had taught him didn’t always work!

SO, TO ANSWER YOUR QUESTION: It works for 11, 12, and 13. And it would work for 22, 33, 44, 55, 66, 77, 88, and 99. Maybe you could argue that it also works for 10: (10^2)=100. (01^2)=001.

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