The digit at unit's place in the number $(13)^{1225} + (11)^{1915} - (23)^{1225}$ is equal to

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Solution

The correct option is B $1$
$(13)^{1225} + (11)^{1915} - (23)^{1225} = (10 + 3)^{1225} + (10 + 1)^{1915} - (20 + 3)^{1225} = 10 k_{1} + 3^{1225} + 10 k_{2} + 1 - 20 k_{3} - 3^{1225} (k_{1}, k_{2}, k_{3} \in Z) = 10 (k_{1} + k_{2} - 2 k_{3}) + 1$

Hence, the unit place digit is $1$

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Finding Remainder of Numbers of the Form a^b

Standard XI Mathematics

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The digit at unit's place in the number (13)1225+(11)1915−(23)1225 is equal to

The correct option is B 1(13)1225+(11)1915−(23)1225=(10+3)1225+(10+1)1915−(20+3)1225=10k1+31225+10k2+1−20k3−31225(k1,k2,k3∈Z)=10(k1+k2−2k3)+1 Hence, the unit place digit is 1

The digit at unit's place in the number $(13)^{1225} + (11)^{1915} - (23)^{1225}$ is equal to

The correct option is B $1$
$(13)^{1225} + (11)^{1915} - (23)^{1225} = (10 + 3)^{1225} + (10 + 1)^{1915} - (20 + 3)^{1225} = 10 k_{1} + 3^{1225} + 10 k_{2} + 1 - 20 k_{3} - 3^{1225} (k_{1}, k_{2}, k_{3} \in Z) = 10 (k_{1} + k_{2} - 2 k_{3}) + 1$

Hence, the unit place digit is $1$