◆Check whether (26)^n can end with the digit 5 for any n € N
◆Check whether 2^n can end with the digit 6 for any n€N
◆Find the units place digit in the expansion of the following -
[tex]({191})^{1009} [/tex]
Kindly give correct ans .
◆Check whether 2^n can end with the digit 6 for any n€N
◆Find the units place digit in the expansion of the following -
[tex]({191})^{1009} [/tex]
Kindly give correct ans .
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_____________________________
1.) Suppose,
n = 0, 1, 2 , 3 , 4
26^0 = 1
26^1 = 26
26^2 = 676
26^3 = 17576
26^4 = 456976
And so on.....
Each time, the last digit us 6 except in the first case , ie - 26^0 = 1.
Thus, 26^n cannot end with 5.
_________________
2.) To end a number with 6, it should have 2 as it's one of the prime factors,
And as 2^n has 2 as it's one of the factors ( 2 × 1)^n
2^n will end with 6 only if it's divisible by 4.
So yes it can end with 6.
_________________
3.)
(191) ^ 1009
We know,
1^n = 1
[ 1 raised to any power always gives 1 ]
So, we conclude, that (191)^1009 will have 1 at its unit place :)
_____________________________
Hope it helps...!!!
Verified answer
Hey mate !!Here's the answer !!
First let me solve one by one.
So taking the first case of
So we know that for a number to end with 5, the prime factorisation of a number must have 5 as it's prime factor.
So calculating prime factorisation of 26 we get,
26 = 13 × 2
Since 26 has no prime factor 5 in it's prime factorisation, there is no natural number ' n ' for which
Second case :
For a number to end with 6, the prime factorisation of that number must be having both 2. But the prime factorisation of 2 is 2. Since it has 2 as it's factor, there is a natural number " n " for which
Third case : Prediction of the unit's digit for the number with expansion
( 191 ) ¹⁰⁰⁹
We know that the number 1 raised to any power gives 1 as the answer in the unit's place. That is,
Here ' n ' is any natural number
Hence we can say that,
( 191 ) ¹⁰⁰⁹ will have 1 at it's unit's place.
Hope my answer helped !!
Cheers !!