check whether 6^n can end with the digit 0 for any natural number n
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Answered by
31
Hey!
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Let us think of some numbers which end the digit 0
10 = 5 × 2
100 = 2 × 2 × 5 × 5
Many more...
Now, Clearly we can see that numbers ending with 0 has prime factors as 2 and 5, take any number ending with digit 0 , it will do have 2 and 5 as a factor.
But,
6 ^ n = (3 × 2)^n
It doesn't have 5 as it's factor thus 6^n cannot end with zero for any natural number n.
_______________
Hope it helps...!!!
_______________
Let us think of some numbers which end the digit 0
10 = 5 × 2
100 = 2 × 2 × 5 × 5
Many more...
Now, Clearly we can see that numbers ending with 0 has prime factors as 2 and 5, take any number ending with digit 0 , it will do have 2 and 5 as a factor.
But,
6 ^ n = (3 × 2)^n
It doesn't have 5 as it's factor thus 6^n cannot end with zero for any natural number n.
_______________
Hope it helps...!!!
Answered by
3
Answer:
Step-by-step explanation:
Any number which ends with zero should have 2 and 5 as it factors
Example 100=2×2×5×5
6^n=(3×2)^n
Because it doesn't have 5 in it's factors it will not end with zero
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