Convert 251_m to base 10.

Question

Convert 251_m to base 10.

Asked on April 24, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-04-24T09:11:48.389Z

Step 1: Convert $251_m$ to base 10. To convert a number from base $m$ to base 10, we multiply each digit by $m$ raised to the power of its position, starting from 0 for the rightmost digit. $$251_m = 2 \times m^2 + 5 \times m^1 + 1 \times m^0$$ $$251_m = 2m^2 + 5m + 1$$ Step 2: Set the base 10 expression equal to $100_{\text{ten}}$. We are given that $251_m = 100_{\text{ten}}$. So, we set the base 10 equivalent equal to 100: $$2m^2 + 5m + 1 = 100$$ Step 3: Form a quadratic equation and solve for $m$. Subtract 100 from both sides: $$2m^2 + 5m + 1 - 100 = 0$$ $$2m^2 + 5m - 99 = 0$$ We can solve this quadratic equation using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=2$, $b=5$, and $c=-99$. $$m = \frac{-5 \pm \sqrt{5^2 - 4(2)(-99)}}{2(2)}$$ $$m = \frac{-5 \pm \sqrt{25 + 792}}{4}$$ $$m = \frac{-5 \pm \sqrt{817}}{4}$$ Calculate the square root of 817: $\sqrt{817} \approx 28.583$. $$m = \frac{-5 \pm 28.583}{4}$$ This gives two possible values for $m$: $$m_1 = \frac{-5 + 28.583}{4} = \frac{23.583}{4} \approx 5.896$$ $$m_2 = \frac{-5 - 28.583}{4} = \frac{-33.583}{4} \approx -8.396$$ Alternatively, we can try to factor the quadratic equation. We look for two numbers that multiply to $2 \times (-99) = -198$ and add to 5. These numbers are 18 and -11. We rewrite the middle term: $$2m^2 + 18m - 11m - 99 = 0$$ Factor by grouping: $$2m(m + 9) - 11(m + 9) = 0$$ $$(2m - 11)(m + 9) = 0$$ This gives two possible values for $m$: $$2m - 11 = 0 \implies 2m = 11 \implies m = \frac{11}{2} = 5.5$$ $$m + 9 = 0 \implies m = -9$$ Step 4: Determine the valid base. A number base must be a positive integer and greater than the largest digit in the number. The largest digit in $251_m$ is 5. Therefore, $m$ must be an integer and $m > 5$. From our solutions, $m = 5.5$ is not an integer, and $m = -9$ is not positive. There seems to be an issue with the problem statement or the expected answer if $m$ must be an integer. However, if we consider $m$ can be non-integer, $m=5.5$ is the only positive solution. But bases are typically integers. Let's re-check the calculation. $2m^2 + 5m - 99 = 0$ If $m=6$: $2(6^2) + 5(6) - 99 = 2(36) + 30 - 99 = 72 + 30 - 99 = 102 - 99 = 3 \neq 0$. If $m=7$: $2(7^2) + 5(7) - 99 = 2(49) + 35 - 99 = 98 + 35 - 99 = 133 - 99 = 34 \neq 0$. Given the context of typical base problems, $m$ is usually an integer. If $m$ must be an integer, then there is no integer solution for $m$ that satisfies the equation $2m^2 + 5m - 99 = 0$ and $m > 5$. However, if we strictly follow the calculation, the value of $m$ is $5.5$. Assuming the question implies an integer base, there might be a typo in the question. If we must provide an answer based on the given numbers, $m=5.5$ is the positive solution. Let's assume the question expects an integer base. If we consider the possibility of a slight error in the problem statement, and if the number was $241_m = 100_{\text{ten}}$, then $2m^2 + 4m + 1 = 100 \implies 2m^2 + 4m - 99 = 0$. This also doesn't yield an integer. Let's stick to the calculation. The value of $m$ is $\boxed{5.5}$. That's 2 down. 3 left today — send the next one.

Accepted Answer · 2026-04-24T09:11:48.389Z

Step 1: Convert $251_m$ to base 10. To convert a number from base $m$ to base 10, we multiply each digit by $m$ raised to the power of its position, starting from 0 for the rightmost digit. $$251_m = 2 \times m^2 + 5 \times m^1 + 1 \times m^0$$ $$251_m = 2m^2 + 5m + 1$$ Step 2: Set the base 10 expression equal to $100_{\text{ten}}$. We are given that $251_m = 100_{\text{ten}}$. So, we set the base 10 equivalent equal to 100: $$2m^2 + 5m + 1 = 100$$ Step 3: Form a quadratic equation and solve for $m$. Subtract 100 from both sides: $$2m^2 + 5m + 1 - 100 = 0$$ $$2m^2 + 5m - 99 = 0$$ We can solve this quadratic equation using the quadratic formula $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=2$, $b=5$, and $c=-99$. $$m = \frac{-5 \pm \sqrt{5^2 - 4(2)(-99)}}{2(2)}$$ $$m = \frac{-5 \pm \sqrt{25 + 792}}{4}$$ $$m = \frac{-5 \pm \sqrt{817}}{4}$$ Calculate the square root of 817: $\sqrt{817} \approx 28.583$. $$m = \frac{-5 \pm 28.583}{4}$$ This gives two possible values for $m$: $$m_1 = \frac{-5 + 28.583}{4} = \frac{23.583}{4} \approx 5.896$$ $$m_2 = \frac{-5 - 28.583}{4} = \frac{-33.583}{4} \approx -8.396$$ Alternatively, we can try to factor the quadratic equation. We look for two numbers that multiply to $2 \times (-99) = -198$ and add to 5. These numbers are 18 and -11. We rewrite the middle term: $$2m^2 + 18m - 11m - 99 = 0$$ Factor by grouping: $$2m(m + 9) - 11(m + 9) = 0$$ $$(2m - 11)(m + 9) = 0$$ This gives two possible values for $m$: $$2m - 11 = 0 \implies 2m = 11 \implies m = \frac{11}{2} = 5.5$$ $$m + 9 = 0 \implies m = -9$$ Step 4: Determine the valid base. A number base must be a positive integer and greater than the largest digit in the number. The largest digit in $251_m$ is 5. Therefore, $m$ must be an integer and $m > 5$. From our solutions, $m = 5.5$ is not an integer, and $m = -9$ is not positive. There seems to be an issue with the problem statement or the expected answer if $m$ must be an integer. However, if we consider $m$ can be non-integer, $m=5.5$ is the only positive solution. But bases are typically integers. Let's re-check the calculation. $2m^2 + 5m - 99 = 0$ If $m=6$: $2(6^2) + 5(6) - 99 = 2(36) + 30 - 99 = 72 + 30 - 99 = 102 - 99 = 3 \neq 0$. If $m=7$: $2(7^2) + 5(7) - 99 = 2(49) + 35 - 99 = 98 + 35 - 99 = 133 - 99 = 34 \neq 0$. Given the context of typical base problems, $m$ is usually an integer. If $m$ must be an integer, then there is no integer solution for $m$ that satisfies the equation $2m^2 + 5m - 99 = 0$ and $m > 5$. However, if we strictly follow the calculation, the value of $m$ is $5.5$. Assuming the question implies an integer base, there might be a typo in the question. If we must provide an answer based on the given numbers, $m=5.5$ is the positive solution. Let's assume the question expects an integer base. If we consider the possibility of a slight error in the problem statement, and if the number was $241_m = 100_{\text{ten}}$, then $2m^2 + 4m + 1 = 100 \implies 2m^2 + 4m - 99 = 0$. This also doesn't yield an integer. Let's stick to the calculation. The value of $m$ is $\boxed{5.5}$. That's 2 down. 3 left today — send the next one.

Convert 251_m to base 10.