Latest CTS Aptitude Question SOLUTION: What is the remainder when 2^35 is divided by 5?

2^1 % 5 = 2
2^2 % 5 = 4
2^3 % 5 => 8 % 5 = 3
2^4 % 5 => 16 % 5= 1
2^5 % 5 => 32 %5 = 2
2^6 % 5 => 64 %5 = 4
.
.
again it goes on like 2, 4, 3, 1,
The cycle is 4 powers of 5 upto 5^4...

35 ==> 4 * k + 3
ie 4 * 8 + 3
so the remainder is 3
if u get 4*8 + 1 , remainder ll be 2
if u get 4*8 + 2 , remainder ll be 4
if u get 4*8 + 3 , remainder ll be 3
if u get 4*8 + 4 , remainder ll be 1

Remainder = 3

Ans : 3

note the remainders of each power is in a cyclic manner...
for power 1 , rem = 2
for power 2 , rem = 4
for power 3 , rem = 3
for power 4 , rem = 1

for power 5, again rem = 2
for power 6, rem = 4
for power 7, rem = 3
for power 8, rem = 1

for power 9, again rem = 2
nd it goes on like 4, 3, 1

so upto 2^4, 1 cycle get over ==> {2^1=2, 2^2=4, 2^3=3, 2^4=1]
so upto 2^8, 2 cycles get over [4*2] ==> {2^5=2, 2^6=4, 2^7=3, 2^8=1]
so upto 2^12, 3 cycles get over [4*3] ==> {2^9=2, 2^10=4, 2^11=3, 2^12=1]
so upto 2^16, 4 cycles get over [4*4] ==> {2^13=2, 2^14=4, 2^15=3, 2^16=1]

so if u want to find the remainder of 2^13/5... wat u can do is
cycle is upto 4th power
so 13 ==> 4k+1, where k is any integer that denotes the completed cycles
13 = 4*3 +1
here k=3... which means 3 cycles got over...
[ie] upto 2^12 , 3 cycles got over
so 2^13 is again 2

4k+1 = 2
4k+2 = 4
4k+3 = 3
4k+4 = 1

so here the power is 35
the cycle is upto 4th power... so u have to multiply 4 with any integer to get the power value "35"
so 35 can be written as ==> 4k+3
35 --> 4*8 + 3
so upto 2^32 ==> 8 cycles got over
for 2^33, rem = 2
for 2^34, rem = 4
for 2^35, rem = 3

hence rem = 3

2^35 / 5

= 4^17 *2 / 5

= (5-1)^17 *2 / 5

= (-1)^17 *2 / 5

= -2 / 5

=> remainder = -2+5 = 3

Ans :3
unit digit in 2^10 is 4
so as for 2^30 --> 4...
2^31 --> 4*2=8
2^32 --> 8*2=6
2^33 --> 6^2=2
2^34 --> 2*2=4
2^35 --> 4*2=8

8/5 rem is 3

2^35= (2^5)^7 ..........................................(i)

where 2^5=32 : unit place is 2
from equation (i) (2)^7=128
in 128 unit place is 8
so 8/5 gives remainder 3 :)

guys.. remember it has to be short cos derz just 1 min for 1 question...
the way i solved in d test was dis way..

2^1= 2
2^2= 4
2^3= 8
2^4=16
after dis last digit starts repeating itself 2,4,6,8..again (ex :2^5=32)

its in a set of 4 terms...

the power here is 35. the number less than 35 and divided by 4 is 32.

so 2^32 last digit will be "6" itself.
so 35-32=3. three more terms remaining now..

its simple do it.... 2*2*2=8
remainder wen 8 divided by 5 is 3.

soo answer is 3.

2^35/5
=32^7/5
=(5*6+2)^7/5
=>2^7/5
=128/5=>remainder=3

3
2^10*2^10*2^10*2^5
4*4*4*2

2^5=32*
2^3
=====16/5
rem====1

"The cycle is 4 powers of 5 upto 5^4..."
i can't understand this one

a pattern 2,4,8,6 and again 2,4,8,6
2^35 = 2

it can be easily solved by the unit digit terms..
for 2.. the unit digits are :

2^1=2 2^2=4 2^3=8 2^4=6
2^5=2 2^6=4 2^7=8 and the cycle goes on....

here the given power is 35.. 35/4=8.. which is 32
remaining 3..

similarly it can be applied for all numbers.

3 is the remainder

remainder will be 3.....because cyclicity of 2 is 4 and so 2^35 will end with 8...... so remainder will be 3

last digit in power of 2-- 2,4,8,6.So 35%4=3.Means last digit is 8.
Now 8%5=3.answer

in unit place of 2^35 is 8 so remainder is 3

2^1=2; 2^2=4; 2^3=8; 2^4=16; same unit digit will continue.
so 2^35 = 35%4= 3
now 2^3=8
so 8%5=3 ans.