Problem Solving

(1941^3843)+(1961^4181)

vigneshraj wrote:(1941^3843)+(1961^4181)

When an integer with an ODD UNITS DIGIT OTHER THAN 5 is raised to A POWER THAT IS A MULTIPLE OF 20, the last two digits of the result are 01.

Thus, the last two digits of 1941³��� and 1961�¹�� are 01.
To determine the last two digits of 1941³��³ and 1961�¹�¹, keep multiplying the last two digits by 41.

Last two digits of 1941³��³:
1941³��� --> 01.
1941³��¹ --> 01*41 = 41.
1941³��² --> 41*41 = 1681.
1941³��³ --> 81*41 = 3321.
Thus, the last two digits of 1941³��³ are 21.

Last two digits of 1961�¹�¹:
1961�¹�� --> 01.
1961�¹�¹ --> 01*61 = 61.
Thus, the last two digits of 1961�¹�¹ are 61.

Thus:
(last two digits of 1941³��³) + (last two digits of 1961�¹�¹) = 21+61 = 82.

Interesting question, but beyond the scope of the GMAT.