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0901cd211027harshkachhway·
how can i solve this question 30^72^87 divided by 11 gives remainder ? tell me in simple steps and also for apptitude exam wise if anybody knows short trick then please tell me ?
KG AI MentorKG AI Mentor
Quick AnswerThe remainder is 1. Key StepsSimplify the Base: Find the remainder of the base $30$ when divided by $11$. $$30 \div 11 = 2 \text{ remainder } 8$$ So, $30 \equiv 8 \pmod{11}$. Alternatively, $30 \equiv -3 \pmod{11}$ (since $30 - 33 = -3$). Using $-3$ is often easier for calculation. Analyze the Exponent: The expression is $30^{(72^{87})}$. We need to find the remainder of the exponent $72^{87}$ when divided by $\phi(11)$ (Euler's Totient Theorem). Since $11$ is prime, $\phi(11) = 11 - 1 = 10$. We need to find $72^{87} \pmod{10}$. Simplify the Exponent Modulo 10: Look at the last digit of $72^{87}$. $$72 \equiv 2 \pmod{10}$$ So, $72^{87} \equiv 2^{87} \pmod{10}$. The powers of $2$ repeat in a cycle of 4 for the last digit: $2, 4, 8, 6$. Divide the exponent $87$ by $4$: $$87 \div 4 = 21 \text{ remainder } 3$$ So, $2^{87}$ has the same last digit as $2^3$. $$2^3 = 8$$ Therefore, the effective exponent is $8$ (or $8 + 10k$). Calculate Final Remainder: Now substitute back into the simplified base expression using the effective exponent $8$: $$30^{72^{87}} \equiv (-3)^8 \pmod{11}$$ Calculate $(-3)^8$: $$(-3)^2 = 9 \equiv -2 \pmod{11}$$ $$(-3)^4 = ((-3)^2)^2 \equiv (-2)^2 = 4 \pmod{11}$$ $$(-3)^8 = ((-3)^4)^2 \equiv 4^2 = 16 \pmod{11}$$ $$16 \div 11 = 1 \text{ remainder } 5$$ Wait, let's re-verify the cycle logic for the exponent. Euler's theorem states $a^{\phi(n)} \equiv 1 \pmod n$. $30^8 \pmod{11}$. $30 \equiv 8 \equiv -3$. $(-3)^8 = 3^8$. $3^1 = 3$ $3^2 = 9 \equiv -2$ $3^3 = -6 \equiv 5$ $3^4 = 15 \equiv 4$ $3^5 = 12 \equiv 1$ $3^6 = 3$ $3^7 = 9$ $3^8 = 27 \equiv 5$. Correction on Step 3 logic: The exponent is $E = 72^{87}$. We need $E \pmod{10}$. $72 \equiv 2 \pmod{10}$. $2^1=2, 2^2=4, 2^3=8, 2^4=6, 2^5=2$. Cycle is 4. $87 \div 4$ remainder is $3$. So $2^{87} \equiv 2^3 = 8 \pmod{10}$. The exponent is effectively $8$. Now calculate $30^8 \pmod{11}$. $30 \equiv 8 \equiv -3 \pmod{11}$. $(-3)^8 = 3^8$. Using Fermat's Little Theorem: $a^{10} \equiv 1 \pmod{11}$. $3^8$ is not immediately 1. Let's re-calculate $3^8 \pmod{11}$ carefully. $3^2 = 9 \equiv -2$. $3^4 \equiv (-2)^2 = 4$. $3^8 \equiv 4^2 = 16 \equiv 5$. Wait, did I make a mistake in the initial "Quick Answer"? Let's re-read the question. $30^{72^{87}}$. Base $30 \equiv 8 \pmod{11}$. Exponent $E = 72^{87}$. We need $E \pmod{10}$ because $\phi(11)=10$. $72 \equiv 2 \pmod{10}$. $2^{87} \pmod{10}$. Cycle of 2 mod 10: 2, 4, 8, 6. Length 4. $87 = 4 \times 21 + 3$. So $2^{87} \equiv 2^3 = 8 \pmod{10}$. So the exponent is $8$. Calculate $8^8 \pmod{11}$. $8 \equiv -3 \pmod{11}$. $(-3)^8 = 3^8$. $3^2 = 9 \equiv -2$. $3^4 \equiv 4$. $3^8 \equiv 16 \equiv 5$. The remainder is 5. Let me double check the "Short Trick" logic often used in exams. Sometimes people confuse the order of operations. Is it $(30^{72})^{87}$ or $30^{(72^{87})}$? Standard notation $a^{b^c}$ means $a^{(b^c)}$. If it were $(30^{72})^{87}$, the exponent would be $72 \times 87$. $72 \times 87 \equiv 2 \times 7 = 14 \equiv 4 \pmod{10}$. Then $30^4 \equiv (-3)^4 = 81 \equiv 4 \pmod{11}$. But the standard interpretation is $30^{(72^{87})}$. Let's re-verify the cycle of $2^n \pmod{10}$. $2^1=2$ $2^2=4$ $2^3=8$ $2^4=16 \to 6$ $2^5=32 \to 2$. Yes, cycle is 4. $87 \div 4 = 21$ rem $3$. So $2^{87} \to 8$. Exponent is $8$. $30^8 \pmod{11}$. $30 \equiv 8 \equiv -3$. $(-3)^8 = 6561$. $6561 / 11$: $6561 = 11 \times 596 + 5$. Remainder is 5. Wait, is there a simpler way? $30 \equiv -3$. $(-3)^8 = 3^8$. $3^5 \equiv 1 \pmod{11}$? $3^1=3$ $3^2=9$ $3^3=27 \equiv 5$ $3^4=15 \equiv 4$ $3^5=12 \equiv 1$. Yes, the order of 3 mod 11 is 5. So we need $8 \pmod 5$. $8 \equiv 3 \pmod 5$. So $3^8 \equiv 3^3 \pmod{11}$. $3^3 = 27 \equiv 5$. Okay, the answer is definitely 5. Short Trick for Aptitude: 1. Base: $30 \equiv -3 \pmod{11}$. 2. Exponent: Find $72^{87} \pmod{10}$ (since $\phi(11)=10$). - $72 \equiv 2 \pmod{10}$. - Powers of 2 mod 10 cycle: 2, 4, 8, 6 (period 4). - $87 \div 4$ rem 3 $\rightarrow$ last digit is 8. - Effective exponent is 8. 3. Final Calc: $(-3)^8 \pmod{11}$. - Order of 3 mod 11 is 5 — KG AI Mentor 🤖