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18 May 2020, 04:38
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Question Stats:
34% (02:12) correct
66%
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Which of the following are factors of \(3^{200} – 5^{100}\)?
I. 7
II. 12
III. 53
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
Are You Up For the Challenge: 700 Level Questions
I. 7
II. 12
III. 53
A. I only
B. II only
C. III only
D. I and III only
E. I, II and III
Are You Up For the Challenge: 700 Level Questions
18 May 2020, 05:32
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\(3^{200} – 5^{100}\)
\(3^{200}\) is a multiple of 3, but \(5^{100}\) is not. Hence their difference can never be a multiple of 3 or 12.
\(9^{100} – 5^{100}\)
\(81^{50} – 25^{50}\)
\(a^n-b^n\) is always a multiple of (a-b) and (a+b), when n is positive even number.
\(81^{50} – 25^{50}\) = (106)*56*k = 2*53*7*8*k
D
\(3^{200}\) is a multiple of 3, but \(5^{100}\) is not. Hence their difference can never be a multiple of 3 or 12.
\(9^{100} – 5^{100}\)
\(81^{50} – 25^{50}\)
\(a^n-b^n\) is always a multiple of (a-b) and (a+b), when n is positive even number.
\(81^{50} – 25^{50}\) = (106)*56*k = 2*53*7*8*k
D
Bunuel
18 May 2020, 05:38
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Let us say, N = 3^200 – 5^100
N = 9^100 – 5^100.
This is a multiple of 9 - 5. Hence, 4, 2 and 1 are factors of the dividend.
N can also be written as 81^50 – 25^50.
This is a multiple of 81 - 25 = 56. Hence, 56, 7, 8 are factors of the dividend.
Dividend = (81^2)25 – (25^2)25
This is a multiple of 81^2 - 25^2, which is equal to (81 - 25) (81 + 25) = 56 × 106 = 7 × 8 × 2 × 53
N = 9^100 – 5^100.
This is a multiple of 9 - 5. Hence, 4, 2 and 1 are factors of the dividend.
N can also be written as 81^50 – 25^50.
This is a multiple of 81 - 25 = 56. Hence, 56, 7, 8 are factors of the dividend.
Dividend = (81^2)25 – (25^2)25
This is a multiple of 81^2 - 25^2, which is equal to (81 - 25) (81 + 25) = 56 × 106 = 7 × 8 × 2 × 53
