 Hello friends, I am Sanjay Gupta. In this video, I am going to demonstrate you how you can check each number of metrics, whether it is Palindrom or not. Before starting, you can note my information. You can follow or subscribe my YouTube channel through the URL youtube.com slash sanjaygupta underscore techool. You can download my programming app Techimus, which is available on Google Play. Now I am going to implement the solution of this problem with the help of C programming. So first time including a header file stdio.h, then I am implementing main function. Inside main, I am declaring a 2D array. Then I J variables, then N, T, S. These variables are declared for calculation purpose. Now with the help of printf, I am going to display the masses, enter elements on console. To receive elements from user, I am applying nested loops. So here I loop will be representing row indexes and J loop will be representing column indexes. With the help of scatter, all elements will be received from user and will be assigned into a matrix. After completion of reading operation, now I have to apply the logical part. So for that purpose, again, I am implementing nested loop. Now inside this loop, I am assigning value of aij into N. So now I have to check whether the value available inside N variable is palantrum or not. So I have initialized S with 0. Now first I am calculating reverse of N. So for that purpose, I am applying the condition N greater than 0. Inside this condition, I have to apply the logic so that I can reverse N. So first statement is T equals to N model is 10, then S equals to S into 10 plus T, then N equals to N by 10. With the help of this logic, reverse of N will be available inside S. So this logic is for reverse. Now after this reverse operation, I have to check S with N. But I can't use N here because after completion of Y loop, value of N will be 0. So in place of N, I can use aij because N is containing the value which is available inside aij. So I can directly use aij in place of N. So if the reverse which is available in S is equals to the original upper which is available inside aij, then we can say number is palantrum. So I can print %d is palantrum, S or you can write aij also here. So S and aij both will be containing same value. So you can use any particular notation whether aij or S here. So this is the complete logic. After completion of all loops, I am writing return 0. So with this logic, I am going to display whether I am going to check whether each number which is available inside matrix is palantrum or not. Now I am going to execute this code. So you can see I am entering different values. Now you can see the output 11, 33, 44, 77, 55, 66. All these numbers are prime remaining 12, 56, 45. These are not prime numbers. So six num sorry palantrum numbers. So six numbers are palantrum and remaining three numbers are not palantrum because their reverse is different from original. So this way I have checked whether the number which is stored inside matrix is palantrum or not. I hope you have understood the logic well. If you want to watch more programming related videos, you can follow or subscribe my YouTube channel through the URL youtube.com. You can download my programming app Techimage which is available on Google Play. Thank you for watching this video.