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Show that 12 n cannot end with zero or 5

WebJul 17, 2014 · 25 n = (5 X 5) n If we raise 5 to any power ,unit digit will only be 5 . For any natural number to end with 0 , it must be a factor of 10 so it must have factors both 2 and 5 .But 25 does not have 2 as a factor. Thus 25 n cannot end with 0 for any natural number n Hope this would have cleared your doubt . WebSolution. If 6 n is end with zero for a natural number n, it should be divisible by 2 and 5. This means that prime factorization of 6 n should contain the prime number 2 and 5. But it is not possible because 6 n=2 n∗3 n so 2 & 3 is only prime in the factorization of 6 n. Since 5 is not present in the prime factorization, there is no natural ...

Show that `12^n` cannot end with the digits `0` or `5` for any …

WebMar 29, 2024 · Let us take the example of a number which ends with the digit 0 So, 10 = 2 5 100 = 2 2 5 5 Here we note that numbers ending with 0 has both 2 and 5 as their prime … fans grinding against radiator pc https://korperharmonie.com

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WebOct 10, 2024 · If 12 n ends with the digit zero it must be divisible by 5. This is possible only if the prime factorisation of 12 n contains the prime number 5. The prime factorisation of 12 … WebNov 27, 2024 · Solution: If the number 12n, for any natural number n, ends with the digit 0 or 5, then it is divisible by 5. That is, the prime factorization of 12n contains the prime 5. This … WebShow that 12n cannot end with the digit 0 or 5 for any natural number n. If any number ends with the digit 0 or 5, it is always divisible by 5. If 12n ends with the digit zero it must be … fan shackle

show that 12 to the power n cannot end with the digit 0 or 5 for any

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Show that 12 n cannot end with zero or 5

Check whether 6^n can end with digit 0 for any natural number n

WebThe prime factors of 12 are 2 and 3. 12 = 2 2 × 3 ⇒ 12 n = 2 2 n × 3 n Since, 5 is not the factor of 12 n Therefore, for any value of n, 12 n will not be divisible by 5. Hence, 12 n cannot end with the digit 0 or 5 for any value of n. Disclaimer: The question is: 12 n cannot end with the digit 0 or 5 for _______ value of n. Suggest Corrections 1 WebSolution. If any number ends with the digit 0 or 5, it is always divisible by 5. If 12 n ends with the digit zero it must be divisible by 5. This is possible only if prime factorisation of 12n …

Show that 12 n cannot end with zero or 5

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WebSolution. If any number ends with the digit 0 or 5. It is always divisible by 5. If 12 n ends with the digit zero or five it must be divisible by 5. This is possible only if prime factorisation of … WebTo Show :- 12 n cannot end with 0 Proof :-12 n = 4 n x 3 m = 2 n x 3 m But the formula of factorisation tells that the numbers can end with 0 or 5 when they are in form of 2 n + 5 m …

WebNov 2, 2024 · To show that 12^n cannot end with 0 or 5 for any natural number n. Explanation: 12 ^n in prime factor form cn be written as, As, there is no term in the factor … WebFor a number to end with 0, the number must have 2 & 5 as factors. 10=2×5 100=2 2×5 2 For a number to end with 5, the number must have 5 as a factor. 30=5×6 3 n×(2 2) m=3 n×2 2m This number dosent have 5 or 5 and 2 as its factors. Therefore the number cant end with 0 o 5. Was this answer helpful? 0 0 Similar questions

WebFor smallish numbers, you could try getting a multiple of 6 = 1 + 5 close to your number, find the number of zeroes for 25 / 6 times that and try to revise your estimate. For example for 156 = 6 ∗ 26. So try 26 ∗ 5 ∗ 5 = 650. 650! has 26 ∗ 5 + 26 + 5 + 1 = 162 zeroes. Since you overshot by 6, try a smaller multiple of 6. WebAug 8, 2024 · Here `" " 12 = 2 xx 2 xx 3` `= 2^(2) xx 3` `rArr" " 12^(n) = (2^(2) xx 3)^(n) = 2^(2n) xx 3^(n)` We know that if any numbers ends with the digits 0 or 5, its is always divisible by 5. But the prime factorisation of `12^(n)` does not contain the prime number 5. Hence, `12^(n)` cannot end with the digits 0 or 5 for any natural number n.

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WebIf 12 n ends with the digit zero or five it should be divisible by 5. It is possible if prime factorisation of 12 n has the prime number 5. 12 = 2 × 2 × 3 = 22 × 3. 12 n = (22 × 3) n = 22 … cornerstone podiatry ayrWebAug 26, 2024 · If any number ends with the digit 0 or 5, it is always divisible by 5. This is possible only if prime factorisation of `12^(n)` contains the prime number 5. Now, `12=2 … cornerstone plymouth used car dealerWebApr 11, 2024 · 4.3K views, 492 likes, 148 loves, 70 comments, 48 shares, Facebook Watch Videos from NET25: Mata ng Agila International April 11, 2024 cornerstone podiatry high point nc