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Bunuel
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
(7+2)^100-(7-2)^100
So resultant will be a multiple of 7.
First number is multiple of 3 and second number is not a multiple of 3 so their difference cannot be a multiple of 3.
Third option I doubt is 53.

Posted from my mobile device­
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Which of the following are factors of \(3^{200}–5^{100}\) ?

I. 7
\(3^{200}–5^{100}= 9^{100} - 5^{100}= (7+2)^{100} - (7-2)^{100}= 7m + 2^{100} - 7n -2^{100}= 7 (m-n)\) (divisible by 7)

II. 12 = 3*4
\(9^{100} - 5^{100}= (4+5)^{100} - 5^{100}= 4a + 5^{100}- 5^{100}= 4a\) (divisible by 4)
\(9^{100} - 5^{100} = 9^{100} - (6-1)^{100}= 9k - 6t -1 = 3(3k-2t) -1\) (not divisible by 3).

III. 53
\(9^{100} - 5^{100}= 81^{50} - 25^{50}= (106-25)^{50} - 25^{50}= 106p\) (divisible by 53).

Answer (D). I and III.
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Bunuel
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

Asked: Which of the following are factors of \(3^{200} – 5^{100}\)?

\(3^{200} – 5^{100} = 81^{50} - 25^{50} = 56*106*k \)
\(56*106 = 2*2*2*7*2*53 = 2^4*7*53\)

I. 7
II. 12
III. 53

IMO D­
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Kinshook

Bunuel
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

Asked: Which of the following are factors of \(3^{200} – 5^{100}\)?

\(3^{200} – 5^{100} = 81^{50} - 25^{50} = 56*106*k \)
\(56*106 = 2*2*2*7*2*53 = 2^4*7*53\)

I. 7
II. 12
III. 53

IMO D

till when do we have to solve this ? if we take 9^100- 5^100
(9+5) (9-5) which shows this a multiple of 7 only.
However, if we further reduce it to 81^50-5^50 - we understand it is a multiple of 7 and 53. How do we understand till when do we have do this

Bunuel­
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nikitamaheshwari

Kinshook

Bunuel
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

Asked: Which of the following are factors of \(3^{200} – 5^{100}\)?

\(3^{200} – 5^{100} = 81^{50} - 25^{50} = 56*106*k \)
\(56*106 = 2*2*2*7*2*53 = 2^4*7*53\)

I. 7
II. 12
III. 53

IMO D

till when do we have to solve this ? if we take 9^100- 5^100
(9+5) (9-5) which shows this a multiple of 7 only.
However, if we further reduce it to 81^50-5^50 - we understand it is a multiple of 7 and 53. How do we understand till when do we have do this

Bunuel

till the N is even. when you keep on dividing N by 2, at one point N will become odd. then (a-b) and (a+b) won't be factor a^n - b^n­
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