Adding Numbers In Base Eight

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Hey guys! Today we're diving into a cool math problem that might seem a little tricky at first, but trust me, it's totally doable! We need to add two numbers, 54 and 67, and give our answer in base eight. Now, I know what some of you might be thinking, "Base eight? What's that?" Don't sweat it! It's just a different way of counting compared to our usual base ten system. Think of it like using a different set of LEGO bricks to build your numbers. In base ten, we use digits 0 through 9. In base eight, we only use digits 0 through 7. Once we hit 8, we 'carry over' just like we do in base ten when we hit 10. So, let's break this down step-by-step. First things first, we need to convert our numbers, 54 and 67, into base eight. Remember, these numbers are given in base ten unless otherwise specified. To convert a base ten number to base eight, we use division with remainder. We keep dividing the number by 8 and record the remainders. The base eight representation is formed by reading the remainders from bottom to top. Let's tackle 54 first. We divide 54 by 8. 54 divided by 8 is 6 with a remainder of 6. Now we divide the quotient, 6, by 8. 6 divided by 8 is 0 with a remainder of 6. Reading the remainders from bottom to top, we get 66 in base eight. So, 54extten=66exteight54_{ ext {ten }} = 66_{ ext {eight }}. Pretty neat, right? Now, let's do the same for 67. Divide 67 by 8. 67 divided by 8 is 8 with a remainder of 3. Next, divide the quotient, 8, by 8. 8 divided by 8 is 1 with a remainder of 0. Finally, divide the quotient, 1, by 8. 1 divided by 8 is 0 with a remainder of 1. Reading the remainders from bottom to top, we get 103 in base eight. So, 67extten=103exteight67_{ ext {ten }} = 103_{ ext {eight }}. Awesome! We've successfully converted both numbers to base eight. Now comes the fun part: adding them together in base eight. We need to calculate 66exteight+103exteight66_{ ext {eight }} + 103_{ ext {eight }}. When we add numbers in any base, we align them by their place value, just like we do in base ten. So, we'll set it up like this: 66 _eight +103 _eight ----- Now, we add column by column, starting from the rightmost digit (the 'ones' place). In the ones column, we have 6 + 3. That equals 9. But remember, we're in base eight, and we can't use the digit 9. So, we need to convert 9 into base eight. How many times does 8 go into 9? It goes in 1 time with a remainder of 1. So, 9 in base ten is equivalent to 11exteight11_{ ext {eight }}. We write down the 1 in the ones column and carry over the other 1 to the next column (the 'eights' place). Now, let's look at the eights column. We have 6 + 0, plus the 1 we carried over. That gives us 7. Since 7 is a valid digit in base eight (it's less than 8), we just write down 7. Finally, we move to the next column (the 'sixty-fours' place). We have only a 1 from the number 103exteight103_{ ext {eight }}. So, we just bring down the 1. Putting it all together, our sum is 171exteight171_{ ext {eight }}. So, adding 66exteight+103exteight66_{ ext {eight }} + 103_{ ext {eight }} gives us 171exteight171_{ ext {eight }}. Let's double-check our work to make sure we got it right. We converted 54 to 66exteight66_{ ext {eight }} and 67 to 103exteight103_{ ext {eight }}. Our addition resulted in 171exteight171_{ ext {eight }}. Let's convert 171exteight171_{ ext {eight }} back to base ten to see if it matches 54+6754 + 67. The place values in base eight are ..., 828^2 (64), 818^1 (8), 808^0 (1). So, 171exteight171_{ ext {eight }} means 1imes82+7imes81+1imes80=1imes64+7imes8+1imes1=64+56+1=121extten1 imes 8^2 + 7 imes 8^1 + 1 imes 8^0 = 1 imes 64 + 7 imes 8 + 1 imes 1 = 64 + 56 + 1 = 121_{ ext {ten }}. Uh oh! It looks like I made a mistake somewhere in my addition explanation. Let's retrace our steps for the addition. It's super common to make little slips, and that's why checking is so important! Our setup was: 66 _eight +103 _eight ----- Ones column: 6 + 3 = 9. In base eight, 9 is 1imes8+11 imes 8 + 1, so we write down 1 and carry over 1. Correct so far. Eights column: 6 + 0 + (carried 1) = 7. We write down 7. Correct. Sixty-fours column: 1. We write down 1. So the sum is indeed 171exteight171_{ ext {eight }}. My conversion check seems to be where the error lies. Let me re-evaluate the conversion of the original numbers. Okay, I found my mistake! When I explained how to convert 54 to base eight, I said 54 divided by 8 is 6 with a remainder of 6. That's correct. Then I said to divide the quotient, 6, by 8, which gives 0 with a remainder of 6. Reading bottom up, that makes 66exteight66_{ ext {eight }}. This is correct. Now, let's re-check the conversion of 67. 67 divided by 8 is 8 with a remainder of 3. This is correct. Then, dividing the quotient 8 by 8 gives 1 with a remainder of 0. This is correct. Finally, dividing the quotient 1 by 8 gives 0 with a remainder of 1. This is correct. Reading the remainders from bottom to top gives 103exteight103_{ ext {eight }}. This is correct. So the conversion of the original numbers is correct. Let's re-check the addition step very carefully. We are adding 66exteight66_{ ext {eight }} and 103exteight103_{ ext {eight }}. $66_{ ext {eight }}$ + $103_{ ext {eight }}$ ----- Rightmost column (ones place): 6+3=96 + 3 = 9. In base eight, 9=1imes8+19 = 1 imes 8 + 1. So, we write down '1' and carry over '1' to the next column. $6 extbf{1}$ $6_{ ext {eight }}$ + $103_{ ext {eight }}$ ----- $1_{ ext {eight }}$ Middle column (eights place): 6+0+(extcarried1)=76 + 0 + ( ext{carried } 1) = 7. Since 7 is a valid digit in base eight, we write down '7'. $66_{ ext {eight }}$ + $103_{ ext {eight }}$ ----- $71_{ ext {eight }}$ Leftmost column (sixty-fours place): We have '1' from 103exteight103_{ ext {eight }}. We just bring it down. $66_{ ext {eight }}$ + $103_{ ext {eight }}$ ----- $171_{ ext {eight }}$ So, the sum 66exteight+103exteight=171exteight66_{ ext {eight }} + 103_{ ext {eight }} = 171_{ ext {eight }}. My previous check calculation was wrong. Let's try the check again. 171exteight171_{ ext {eight }} means 1imes82+7imes81+1imes80=1imes64+7imes8+1imes1=64+56+1=121extten1 imes 8^2 + 7 imes 8^1 + 1 imes 8^0 = 1 imes 64 + 7 imes 8 + 1 imes 1 = 64 + 56 + 1 = 121_{ ext {ten }}. The sum of the original numbers in base ten is 54+67=121extten54 + 67 = 121_{ ext {ten }}. Success! The base ten sum matches the base ten conversion of our base eight sum. This means our answer 171exteight171_{ ext {eight }} is correct! Let's quickly look at the options provided: A. 111exteight111_{ ext {eight }}, B. 121exteight121_{ ext {eight }}, C. 123exteight123_{ ext {eight }}, D. 143exteight143_{ ext {eight }}. My result is 171exteight171_{ ext {eight }}, which is not among the options. This indicates a potential misinterpretation of the original numbers or a typo in the question or options. However, if the question intended for the numbers 54 and 67 to already be in base eight, the problem would be very different and much simpler. Let's explore that possibility, as it's a common way these problems are sometimes phrased ambiguously. If 54exteight54_{ ext {eight }} and 67exteight67_{ ext {eight }} are the numbers to be added: $54_{ ext {eight }}$ + $67_{ ext {eight }}$ ----- Rightmost column (eights to the power of 0): 4+7=114 + 7 = 11. In base eight, 11=1imes8+311 = 1 imes 8 + 3. So, we write down '3' and carry over '1'. $5 extbf{1}$ $4_{ ext {eight }}$ + $67_{ ext {eight }}$ ----- $3_{ ext {eight }}$ Leftmost column (eights to the power of 1): 5+6+(extcarried1)=125 + 6 + ( ext{carried } 1) = 12. In base eight, 12=1imes8+412 = 1 imes 8 + 4. So, we write down '4' and carry over '1'. $54_{ ext {eight }}$ + $6 extbf{1}$ $7_{ ext {eight }}$ ----- $43_{ ext {eight }}$ Then we bring down the carried '1'. $54_{ ext {eight }}$ + $67_{ ext {eight }}$ ----- $143_{ ext {eight }}$ So, if the original numbers were in base eight, the sum would be 143exteight143_{ ext {eight }}. Let's check the options again: A. 111exteight111_{ ext {eight }}, B. 121exteight121_{ ext {eight }}, C. 123exteight123_{ ext {eight }}, D. 143exteight143_{ ext {eight }}. Bingo! Option D matches our result if we assume the initial numbers were in base eight. Given that one of the options directly matches this interpretation, it's highly probable that the question meant 'Add 54exteight54_{ ext {eight }} and 67exteight67_{ ext {eight }}, giving the answer in base eight.' This is a common way these problems are presented in textbooks and tests. When no base is specified, it's typically assumed to be base ten. However, in the context of a question about base eight, and with options provided, interpreting the input numbers as base eight is often the intended path. So, the correct answer, based on the provided options, is likely D. 143exteight143_{ ext {eight }}. It's a great reminder, guys, to always pay close attention to the context and the given options. Sometimes, the ambiguity is intentional to test your understanding of different interpretations! Keep practicing, and you'll get the hang of these different number bases in no time!