TCS
Company
Numerical Ability
Number System
what is the remainder when 2^39 is divided by 39?
Read Solution (Total 12)
-
- remainder will be 8.
2^39 = (64)^6 * 8 /39
25 ^ 6 * 8 /39
(625)^3 *8 /39
{1 ^3} * 8 /39
=> 8 - 11 years agoHelpfull: Yes(23) No(6)
- 2^12/39=4096/39 rem=1
2^39=(2^12)^3*(2^3)/39 rem=(1^3)*(2^3)=8
thus ans=8 - 11 years agoHelpfull: Yes(9) No(0)
=(2^39)/39
=[((2^6)^6)*2^3]/39
Apply Remainder the.. 64/39=25
=(25^6)*2^3/39
.
.
.
.
=8
- 11 years agoHelpfull: Yes(4) No(2)
- 2^39 will be written as 2^7*2^7*2^7*2^7*2^7*2^4
when 2^7 divided by 39 leaves a remainder of 11
so 11*11*11*11*11*16(when 2^4 divided by 3 leaves a remainder of 16)
it can be written as 121*121*176
large numbers so can be simplified as
when 121 divided by 39 leaves remainder=4
when 176 divided by 39 leaves remainder=20
so 4*4*20=320
320 divided by 39 leaves remainder=8
Ans:8 - 11 years agoHelpfull: Yes(3) No(1)
- ans : 2
(2^38*2^1)/39
rem: (1*2)/39 // using fermet theorem
=2 - 11 years agoHelpfull: Yes(2) No(5)
- 8
see wolframalpha.com to solve - 11 years agoHelpfull: Yes(1) No(1)
- 4)39(9
36
------
3
2^3=8
remainder=8 - 10 years agoHelpfull: Yes(1) No(1)
- Ans: 2 because (a^p)/p =a
- 8 years agoHelpfull: Yes(1) No(0)
- 2^39=2^6*2^6*2^6*2^6*2^6*2^6*2^3/39
now using remainder theoram we get 25+25+25+25+25+25+8/39=2
so remainder is 2
- 11 years agoHelpfull: Yes(0) No(3)
- reminder will be 8.
2^39/39=((2^6)^6)*2^3/39
so, 2^6/39= reminder will be 25
so, (25^6)*2^3/39
25^6/39= (25^2)^3/39= reminder 1
so, (1^6)*2^3/39= reminder 39 - 11 years agoHelpfull: Yes(0) No(4)
- reminder will be 8.
2^39/39=((2^6)^6)*2^3/39
so, 2^6/39= reminder will be 25
so, (25^6)*2^3/39
25^6/39= (25^2)^3/39= reminder 1
so, (1^6)*2^3/39= 1*8/39= reminder 8 - 11 years agoHelpfull: Yes(0) No(2)
- answer is 2 ......
- 10 years agoHelpfull: Yes(0) No(2)
TCS Other Question