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Consider the equation (213)6 = (a3)b with a & b values to be unknown. Then what are the number of possible solutions for the values a & b?
Correct answer is '4'. Can you explain this answer?
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Most Upvoted Answer
Consider the equation (213)6 = (a3)b with a & b values to be unkno...
Given,
(213)6 = (a3)b
Converting the both in decimal base to form an equation, we get
⇒ 2 × 6 × 6 + 1 × 6 + 3 = ab + 3
⇒ 72 + 6 + 3 = ab + 3
⇒ ab = 78
Now, by finding the factors of 78, we get
= 2 × 3 × 13
Now,
We have two equation which should be followed while deciding the pairs of a & b:
(I) a < b
(II) b > 3
So, the possible solutions for the answer is only 4 l.e.
= (1, 78), (2, 39), (3, 26), (6, 13)
Hence, the correct answer is 4.
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Community Answer
Consider the equation (213)6 = (a3)b with a & b values to be unkno...
In order to determine the value of a, we need to convert the base 6 number (213)6 into base 10.
To do this, we can use the place value system where each digit is multiplied by the corresponding power of the base.
In base 6, the rightmost digit has a place value of 6^0, the second digit from the right has a place value of 6^1, and the third digit from the right has a place value of 6^2.
So, (213)6 = 2 * 6^2 + 1 * 6^1 + 3 * 6^0
= 2 * 36 + 1 * 6 + 3 * 1
= 72 + 6 + 3
= 81
Therefore, (213)6 = 81.
Now, we can substitute this value into the equation (a3)b = 81.
(a3)b = 81
(a * b^2) + (3 * b^0) = 81
(a * b^2) + 3 = 81
a * b^2 = 81 - 3
a * b^2 = 78
Since we don't have any information about the base b, we cannot determine the exact value of a.
To do this, we can use the place value system where each digit is multiplied by the corresponding power of the base.
In base 6, the rightmost digit has a place value of 6^0, the second digit from the right has a place value of 6^1, and the third digit from the right has a place value of 6^2.
So, (213)6 = 2 * 6^2 + 1 * 6^1 + 3 * 6^0
= 2 * 36 + 1 * 6 + 3 * 1
= 72 + 6 + 3
= 81
Therefore, (213)6 = 81.
Now, we can substitute this value into the equation (a3)b = 81.
(a3)b = 81
(a * b^2) + (3 * b^0) = 81
(a * b^2) + 3 = 81
a * b^2 = 81 - 3
a * b^2 = 78
Since we don't have any information about the base b, we cannot determine the exact value of a.
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Consider the equation (213)6 = (a3)b with a & b values to be unknown. Then what are the number of possible solutions for the values a & b?Correct answer is '4'. Can you explain this answer?
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