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the last two digitss in 1941^3483+1961^4181=
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- rule : [any odd no's not ending with 5]^20m thn its last digit will be 01
1941^3480 *1941^3+ 1961^4180*1961= 01*1941^3+01*1961
1941^3=21 as last two digit and when add with 1961 we get 82 as last two digit so 82 is the last two digit.................. - 11 years agoHelpfull: Yes(17) No(6)
- 10's digit i,e 4 multiply by 3 and units digit 1 power of 3 that equal to 21.
6*1,1power 1 =61.61+21=82 - 8 years agoHelpfull: Yes(0) No(0)
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