Problem Solving

Please, could you help me with this? Thanks in advance!!!

If the units digit of n^33 is 7, which of the following could be the value for n?

I. n=41
II. n=43
III. n=47

a. Only I
b. Only II
c. Only III
d. I and II
e. II and III

is OA C?

answer should be Only III

condition one : if n=41 raised to nething .. unit digit is always one

condition two : if n=43 the power raised follows with unit digits sequence of 3,9,7,1,3,9,7,1,3,9,7,1,.......

so on 33 rd term will be 3 as the unit digit

condition two : if n=47 the power raised follows with unit digits sequence of 7,9,3,1,7,9,3,1,7,9,3,1,........

so on 33 rd term will be 7 as the unit digit


So OA is C

Vishu

imo C
I would like to explain it with the method of Cylicity.

Cyclicity is every no when raised to some power..it starts repeating itself.Like 2^1 =2
2^2 =4
2^3 =8
2^4 =16
2^5 = 32 ...Here unit digit is again 2..So cyclicity of 2 is 4 means every 4th power of 2 will have 2 as unit digit. like if 2^245 ..unit digit will be 245/4 = 1as reminder and 2^1 = 2 will be unit digit.
Same is with cyclicty of every no. like
No. Cylicity
2 4
3 4
5 1( 5^1 =5 ,5^2 =25..unit digit 5
7 4


So,In this ques...n^33 =7.
Putting all the values of n and find the last digit..
Like if u put n=41..but obvious last digit = 1.
n = 43 , 43^33..here we have to see the cycliclty of 3^33..so its 4 , 33/4 =reminder 1 , 3^1 =3...hence no...

n=47 ,47^33 , we have to see the cylicity of 7^33 ...as wee know cycliciyt of 7 is 4, 33/4 =reminder 1. so 7^1 =7...Hence ...yes


This is the shortest method to find the unit digit of any no. raised to any power...U Just need to keep in mind the Cyclicity

Any confusion...? You can post

stop@800 wrote:is OA C?

Sorry I forgot, Yes, the OA is C