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. 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 …
Show that 12 n cannot end with zero or 5
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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 …
WebShow that 12 n cannot end with the digit 0 or 5 for any natural number n. Answers (1) We know that, if any number ends up with digit 0 or 5 then it must be divisible by 5. Here 12n … WebApr 25, 2024 · The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is (A) 13 (B) 65 (C) 875 (D) 1750 Solution: (A) Since 5 and 8 are the remainders of 70 and 125, respectively. Thus, after subtracting these remainders from the numbers, we have the numbers (70 – 5) = 65, (125 – 8) = 117 which are divisible by the required number.
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 = … Webnews presenter, entertainment 2.9K views, 17 likes, 16 loves, 62 comments, 6 shares, Facebook Watch Videos from GBN Grenada Broadcasting Network: GBN News 12th April 2024 Anchor: Stasia Blake GBN...
WebIf any number ends with the digit 0 that means it should be divisible by 5. That is, if 6 n ends with the digit 0, then the prime factorisation of 6 n would contain the prime number 5. Prime factors of 6 n = (2 × 3) n = (2) n × (3) n. We can clearly observe, 5 is not present in the prime factors of 6 n. That means 6 n will not be divisible by 5.
WebJan 18, 2024 · See answer. 12 to the power n cannot end with the digit 0 or 5 because for a number to end with the digit 0 or 5,it should have 2 and 5 both in its prime factorization. … milwaukee bucks drag showWebShow 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 … milwaukee bucks derrick roseWebAug 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] milwaukee bucks cream city uniformsWebSolution. 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 … milwaukee bucks decorationsWebThe 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 … milwaukee bucks eric gordonWebApr 21, 2024 · For a number to be end with digit zero its prime factors must have atleast one 2 and A are 5 both and for a number to be end with 5 its prime factors must have at least … milwaukee bucks donation request formWebShow 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 … milwaukee bucks did they win