WebAug 23, 2024 · Best answer 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 divisible by 5. This is possible only if prime factorisation of 12n contains the prime number 5. 12 = 2 × 2 × 3 = 22 × 3 ⇒ 12n = (22 × 3)n = 22n × 3n Since its prime factorisation does not contain 5. Web$\begingroup$ @Git: Of course induction is not impossible to avoid: all you need is some simple axioms about summations, and Josue shows how to make the proof. You can, of course, add induction to anything you want (you used plus? you must have inducted over the successor function!), but that doesn't mean induction was there in the original argument. …
Show that the number 9n cannot … Homework Help myCBSEguide
WebIt 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 12 n contains the prime number 5. Now, 12 = 2 × … WebFeb 10, 2015 · Another way to look at this is: take any six-digit number in which the digits are increasing from left to right. Insert the three digits that were not used, keeping the digits in increasing sequence. For each possible starting number there is only one way to do this; for example, 235678 ↦ 1 23 4 5678 9 ↦ 1 23 4 5678 9. The process is reversible. ios 用 microsoft office アプリ
2. Show that 9n cannot end with digit 0 for any natural number
WebIf 9 n ends with the digit zero or five it must be divisible by 5. This is possible only if prime factorisation of 9 n contains the prime number 5. Now, 9 = 3 × 3 9 n = (3 × 3) n = 3 n × 3 n Since, there is no term containing 5 Therefore, there is no value of n ∈ N for which 9 n ends with the digit zero or five 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 … WebFeb 6, 2024 · Show that 9n cannot end with digit 0 for any natural number n. #a4sdoubtsengine ontrack aftership