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vigneshraj
- Newbie | Next Rank: 10 Posts
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- Joined: Mon Sep 24, 2012 3:58 pm
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.vigneshraj wrote:(1941^3843)+(1961^4181)
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.
