Latest CAT Aptitude Question SOLUTION: Q. What is last two digit of 73^435?

• Placement Papers
• /
• CAT

CAT Question Papers:

(#M40031054) CAT question num sys Keep an EYE

Q. What is last two digit of 73^435?

Asked In CAT Vijay Singh Adhikari (10 years ago)

Unsolved Read Solution (2)

Is this Puzzle helpful?   (1)   (0) Submit Your Solution

Read Solution (2) :

shakshi jajoo  (10 years ago)

73^435 = (73^2)^217 * 73
=(27^2)^217 * 73 {since last two digits of x^2= (100-x)^2 = (50-x)^2 = (50+x)^2}
=(_29*_29)^217 * 73
=(_41)^217 *73
=(_81) * 73
= _13.
{since for numbers ending in 1, last digit is always 1 and the second last digit is product of tens digit of base* the unit digit if power, i.e 4*7=28 in this case)

Is this solution Helpfull? Yes | No   

Umesh Yadav  (10 years ago)

since remainer=435%4=3
therefore, 73^power=___17
So last two digit will be 17

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.