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(2183)^215 find 10s place digit? how do you solve these questions please...?
Brother...i can help u with the unit digit place...!! see...for these kinda sums u need to know cyclicity....like in this case 2183...we consider the last digit..3..now cyclicity of 3 is 4...how?? welll......we know3^1 = 33^2 = 93^3 = 273^4 = 813^5 =24(3)Just check out....3 comes again after 4 times...so the cyclicity becomes 4....hope u get it..NOW,3^215 = 3^212 * 3^3 = 3^3 =2(7)sooo....the last digit is 7!!! hope u get that!!!
I got the answer as 8...........Whats the correct ans?
Yes, the correct answer is 8 but kindly tell the method..
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