TCS
Company
Numerical Ability
Number System
2. Find last two digits of : (1021^3921)+ (3087^3921)
Read Solution (Total 4)
-
- 08 is ans
for 1021^3921 -> 2*1=1(second last digit) and last is 1 so 21
for 3087^3921 -> 87^3920*87 -> 69^1460 *87-> 61^730 * 87 -> 01 *87-> 87
now 21+87= 108
08 last two digit
- 11 years agoHelpfull: Yes(14) No(0)
- 8 is the last digit
- 11 years agoHelpfull: Yes(4) No(6)
- Answer will be 8
- 11 years agoHelpfull: Yes(2) No(5)
- prabhjot singh.....will u plz xplain ur ans
- 10 years agoHelpfull: Yes(0) No(0)
TCS Other Question