Grades 11-12 Video Solutions 2024-2024_11-12

Video Transcription

A student started with the number 1 and multiplied it by either 6 or 10. He then multiplied the result by either 6 or 10 and continued this procedure many times.  Which of the following cannot be one of the numbers he obtained?  Let A be the number of factors of 6 and let B be the number of factors of 10. Note that A and B are both non-negative.  We know that our total result is 6 to the A times 10 to the B which can be rewritten as 2 to the A plus B times 3 to the A times 5 to the B.  Also the power of 2 must be equal to the sum of the powers of 3 and 5. Only answer choice B does not satisfy this condition since 90 is clearly not 100.

Video Summary

A student multiplies the number 1 by either 6 or 10 repeatedly. The results can be expressed in terms of their prime factors: \(6^A \times 10^B\), which equals \(2^{A+B} \times 3^A \times 5^B\). The requirement is that the exponent of 2 (A+B) equals the sum of the exponents of 3 and 5 (A+B). Choice B, which is 90, violates this condition because its prime factorization results do not satisfy this equation (it factors into \(2^1 \times 3^2 \times 5^1\), making the left side 1 not equal to the right side 3).

Keywords

prime factorization

multiplication

exponents

condition violation

number theory