 Hello friends, so we are going to solve another problem and this is related to the topic of fundamental theorem of arithmetic and this is also Borrowed from ncrt book. So this is part of Your board exam preparation. So the question says prove that there is no natural number n no natural number n for which 4 to the power n ends with the digit 0 that means You can't find any natural number n when which when raised to the power on 4 and with digit 0 So now we know anything which ends with this is 0 must be divisor by 5 So let's say if you have 10 so we know that it is 2 into 5 If it is 30 then we know that this is 6 into 5 or 2 into 3 into 5. So one Factor which must be there if there is a number which ends with 0 is 5 so 5 must be there in the as a factor of that particular number now Let us take up 4 to the power n. So 4 to the power n can be written as 2 to the power 2 to the power n Because 4 is 2 square, which is nothing but 2 to the power 2 n Okay, so any number let's say this this number 4 to the power n Let's say a is equal to 4 to the power n. So hence now we can say a is equal to 2 to the power 2n So there is only one prime factor. So only One prime factor is there one prime factor To is there is there yeah for for 2 to the power 2n Okay, so there is no, you know, why is there no 5 then by fundamental theorem of arithmetic Arithmetic we know that we know that any any composite number any composite number will have any composite number any composite number Will have unique Prime factorization unique prime factorization Fine factorization unique prime factorization. What does I what does it mean? It means that if you have expressed any composite number in terms of its fine factor You cannot really find anything beyond that. So all the prime factors will be x it will be there in that Expression right so here here we say that the only prime factor available is to there is no Five in it. So there is no chance that there, you know, this number would be dispelled by five So if this div is this num this number doesn't carry any prime factor As five then there is no possibility of having zero at its Units place. So that is what the proof would look like and This is a very good application of fundamental theorem of arithmetic. See you in the next problem-solving video. Thanks a lot