Show that 12 n cannot end with zero or 5
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. 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 …
Show that 12 n cannot end with zero or 5
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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. WebIf 12 n ends with the digit zero it must be divisible by 5. This is possible only if prime factorisation of 12^n contains the prime number 5. Now, 12 = 2 × 2 × 3 = 2 2 × 3 ⇒ 12 n = …
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 …
WebAug 26, 2024 · Best answer If any number ends with the digit 0 or 5, it is always divisible by 5. This is possible only if prime factorisation of 12n 12 n contains the prime number 5. Now, 12 = 2 × 2 × 3 = 22 × 3 12 = 2 × 2 × 3 = 2 2 × 3 ⇒ 12m = (22 × 3)n = 22n × 3m ⇒ 12 m = ( 2 2 × 3) n = 2 2 n × 3 m [Since, there is no term contains 5] 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 …
Web12 = 2 2 × 3 ∴ 12 n = ( 2 2 × 3) n = ( 2 2) n × 3 n So, only primes in the factorisation of 12 n are 2 and 3 and, not 5. Hence, 12 n cannot end with digit 0 or 5. 0Thank You ANSWER Related Questions Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. Posted by _Jass_ Mahey_ 1 day, 6 hours ago CBSE > Class 10 > Mathematics
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 … bandula jayasinghe visit visa canadaWebJun 8, 2015 · 12n can only end with digits 0 or 5 when it base prime factors include 2 and 5 as its factors. but 12 has not 2 and 5 as its prime factors . so it is impossible for 12n to end with the digit 0 or 5 Advertisement Still have questions? Find … bandulan sukun malangWebJan 17, 2024 · Jan 16, 2024 125 Dislike Share Save Kwatra Tuition Center 18.4K subscribers 8. Show that 12n cannot end with the digit 0 or 5 for any natural number n. Search for #REALNUMBERSBYKTC For... artus sanierung facebookWebShow that (12) n cannot end with digit 0 or 5 for any natural number n. Answers (1) There is no factor of the form 5 n, therefore, 12 n can not end with digit 0 or 5 for any natural … bandulan 2d togelWebMar 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 factors Whereas 4n = (2 × 2) n Does not have 5 as a prime factor. So, it does not end with zero. Therefore, 4n cannot end with zero for any natural number n artus rekadayaWebJan 18, 2024 · Prove that 12^n cannot end with digit 0 or 5 for any natural number nClass 10 Mathematics Chapter 1 Real Numbers Important Question. bandula kumaraWebSolution. 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 ... bandula padmakumara