Linear Congruence Quiz: Solve Modular Equations

The last digit of powers of 3 repeats in the cycle 3, 9, 7, 1 with period 4. To find the last digit of 3 to the power of 47, divide the exponent by 4. 47 divided by 4 gives quotient 11 remainder 3. A remainder of 3 means the exponent corresponds to the 3rd position in the cycle. The 3rd value in the cycle 3, 9, 7, 1 is 7. So the last digit of 3 to the power of 47 is 7.

The last digit of powers of 4 repeats in the cycle 4, 6 with period 2. Odd exponents give last digit 4 and even exponents give last digit 6. Since 53 is odd, 4 to the power of 53 ends in 4. Confirming: 4 to the 1st equals 4, 4 to the 2nd equals 16, 4 to the 3rd equals 64, and the pattern of 4 for odd exponents and 6 for even exponents continues indefinitely.

The answer is True. The number 6 multiplied by any whole number always produces a result ending in 6. For example, 6 to the 1st equals 6, 6 to the 2nd equals 36, 6 to the 3rd equals 216, and 6 to the 4th equals 1296. In modular terms, 6 is congruent to 6 mod 10, and 6 multiplied by 6 equals 36, which is also congruent to 6 mod 10. This pattern holds for all positive exponents without exception.

Since 5 is congruent to 1 modulo 4 (because 5 minus 4 equals 1), raising 5 to any power raises 1 to that same power in modular arithmetic. 1 raised to any power equals 1. Therefore 5 to the power of 37 is congruent to 1 to the power of 37, which equals 1 mod 4. This approach works whenever the base is congruent to 1 modulo the divisor.

Since 11 is congruent to 5 modulo 6 (because 11 minus 6 equals 5), compute powers of 5 mod 6 instead. The cycle of 5 mod 6 is 5, 1 with period 2: 5 to the 1st equals 5, 5 to the 2nd equals 25 which is congruent to 1. Divide the exponent by 2: 35 mod 2 equals 1. The 1st term in the cycle is 5. So 11 to the power of 35 is congruent to 5 mod 6.

Since 7 is congruent to 1 modulo 3 (because 7 minus 6 equals 1), raising 7 to any power raises 1 to that same power in modular arithmetic. 1 raised to any power equals 1. Therefore 7 to the power of 8 is congruent to 1 to the power of 8, which equals 1 mod 3. Confirming: 7 to the 8th equals 5764801; 5764801 divided by 3 gives remainder 1.

First find 3 to the power of 10 mod 5. The cycle of 3 mod 5 is 3, 4, 2, 1 with period 4. 10 divided by 4 gives remainder 2, so the answer corresponds to the 2nd term in the cycle, which is 4. So 3 to the 10th is congruent to 4 mod 5. Now check each option: 9 divided by 5 gives remainder 4 — congruent. 14 divided by 5 gives remainder 4 — congruent. 18 divided by 5 gives remainder 3 — not congruent. 24 divided by 5 gives remainder 4 — congruent. Options A, B, and D all satisfy the condition.

Since 14 is congruent to 5 modulo 9 (because 14 minus 9 equals 5), compute powers of 5 mod 9 instead. The cycle of 5 mod 9 is 5, 7, 8, 4, 2, 1 with period 6. Divide the exponent by 6: 20 mod 6 equals 2. The 2nd term in the cycle is 7. So 14 to the power of 20 is congruent to 7 mod 9.

The answer is False. Since 15 is congruent to 0 modulo 5 (because 15 is exactly divisible by 5), raising 15 to any positive power raises 0 to that same power in modular arithmetic. 0 raised to any positive power equals 0. Therefore 15 to the power of n is congruent to 0 mod 5 for all positive integers n, not 5. The remainder is 0, not 5, because 5 itself is not a valid remainder when dividing by 5.

Since 6 is congruent to negative 1 modulo 7 (because 6 plus 1 equals 7), raising 6 to any power raises negative 1 to that same power in modular arithmetic. Negative 1 raised to an odd power equals negative 1. Since 17 is odd, 6 to the power of 17 is congruent to negative 1 mod 7. Converting negative 1 to a standard remainder: negative 1 plus 7 equals 6. So 6 to the power of 17 is congruent to 6 mod 7.

The cycle of 2 mod 7 is 2, 4, 1 with period 3. Divide the exponent by 3: 31 mod 3 equals 1. The 1st term in the cycle is 2. So 2 to the power of 31 is congruent to 2 mod 7. Confirming the cycle: 2 to the 1st equals 2, 2 to the 2nd equals 4, 2 to the 3rd equals 8 which is congruent to 1 mod 7, then the cycle repeats from 2 again.

Since 10 is congruent to 1 modulo 3 (because 10 minus 9 equals 1), raising 10 to any power raises 1 to that same power in modular arithmetic. 1 raised to any power equals 1. Therefore 10 to the power of 100 is congruent to 1 mod 3. This result holds for any positive power of 10, since 10 is always congruent to 1 modulo 3.

The answer is True. The last digit of a product depends only on the last digits of the factors. Since 12 and 2 have the same last digit, their powers will always produce the same last digit at each step. In modular terms, 12 is congruent to 2 mod 10, so 12 to the n is congruent to 2 to the n mod 10 for all positive integers n. For example, 12 to the 3rd equals 1728 ending in 8, and 2 to the 3rd equals 8, both ending in 8.

Since 17 is congruent to 1 modulo 4 (because 17 minus 16 equals 1), raising 17 to any power raises 1 to that same power in modular arithmetic. 1 raised to any power equals 1. Therefore 17 to the power of 50 is congruent to 1 to the power of 50, which equals 1 mod 4. This approach applies whenever the base leaves remainder 1 when divided by the modulus.

First find 2 to the power of 200 mod 7. The cycle of 2 mod 7 is 2, 4, 1 with period 3. 200 divided by 3 gives remainder 2. The 2nd term in the cycle is 4. So 2 to the 200th is congruent to 4 mod 7. Now check each option: 11 divided by 7 gives remainder 4 — congruent. 15 divided by 7 gives remainder 1 — not congruent. 18 divided by 7 gives remainder 4 — congruent. 25 divided by 7 gives remainder 4 — congruent. Options A, C, and D all satisfy the condition.

The last digit of powers of 8 repeats in the cycle 8, 4, 2, 6 with period 4. To find the last digit of 8 to the power of 47, divide the exponent by 4. 47 divided by 4 gives quotient 11 remainder 3. A remainder of 3 means the exponent corresponds to the 3rd position in the cycle. The 3rd value in the cycle 8, 4, 2, 6 is 2. So the last digit of 8 to the power of 47 is 2.

Since 13 is congruent to 1 modulo 12 (because 13 minus 12 equals 1), raising 13 to any power raises 1 to that same power in modular arithmetic. 1 raised to any power equals 1. Therefore 13 to the power of 30 is congruent to 1 to the power of 30, which equals 1 mod 12. This is a useful shortcut: any number that is exactly 1 more than the modulus will always give remainder 1 when raised to any power.

The answer is False. When squaring both sides of a congruence, the result must be reduced modulo the given divisor. While a squared is congruent to 3 squared equals 9 mod 7, the value 9 must then be reduced mod 7. Since 9 minus 7 equals 2, the correct result is that a squared is congruent to 2 mod 7, not 9. The remainder must always be expressed as a value between 0 and the modulus minus 1.

Since 6 is congruent to 1 modulo 5 (because 6 minus 5 equals 1), raising 6 to any power raises 1 to that same power in modular arithmetic. 1 raised to any power equals 1. Therefore 6 to the power of 25 is congruent to 1 to the power of 25, which equals 1 mod 5. Confirming: 6 to the 1st equals 6, remainder 1 when divided by 5. 6 to the 2nd equals 36, remainder 1 when divided by 5. The pattern continues for all powers.

Computing directly: 5 to the 1st equals 5. 5 to the 2nd equals 25; 25 minus 22 equals 3, so 5 squared is congruent to 3 mod 11. 5 to the 3rd equals 5 multiplied by 3 equals 15; 15 minus 11 equals 4, so 5 cubed is congruent to 4 mod 11. 5 to the 4th equals 5 multiplied by 4 equals 20; 20 minus 11 equals 9, so 5 to the 4th is congruent to 9 mod 11. The answer is 9.