Question

HEYA...... RYTHME HERE
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#NEED FULL EXPLANATION
#10 POINTS
#NOSPAMS

Answer 1

Hint :

(i) 2^10 ends in 24.

(ii) 24 raised to even powers ends in 76 and 24 raised to odd powers ends in 24.

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Given : (2)^999

= > (2^10)^99 * (2)^9

= > (24)^99 * (2)^9

= > 24 * 512

= > 12288.

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Therefore, the last two digits of (2)^999 = 88.

Hope this helps!