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If a, b, and c are positive consecutive odd integers, what is the

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Bunuel
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If a, b, and c are positive consecutive odd integers, what is the [#permalink] New post   06 Oct 2022, 01:30
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If a, b, c and d are consecutive odd integers such that \(123 < a < b < c < d\), then what is the sum of the distinct prime factors of \(2^a + 2^b + 2^c + 2^d\) ?

A. 17
B. 19
C. 22
D. 24
E. 25




 


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Re: If a, b, and c are positive consecutive odd integers, what is the [#permalink] New post   06 Oct 2022, 19:09
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*Since a>123 => 2^a>1
*Since a, b, c, d are consecutive odd numbers
=> b=a+2; c=a+4; d=a+6
=> 2^a+2^b+2^c+2^d=2^a+(2^a*2^2)+(2^a*2^4)+(2^a*2^6)
= 2^a(1+2^2+2^4+2^6)
*Since the question is asking for "distinct prime factors":
(1) 2^a has the prime factor 2
(2) 1+2^2+2^4+2^6 = 1+4+16+64 = 85 = 17*5
(1)(2) => 17+5+2 = 24.
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Re: If a, b, and c are positive consecutive odd integers, what is the [#permalink] New post   06 Oct 2022, 10:24
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Answer should be 24. The odd consecutive factors are 123, 125, 127 and 129. Plug them into the equation. You have 2^123 + 2^125 + 2^127+2^129. Factor out 2^123 and you will then have (1+2^2+2^4+2^6) in the equation. Sum those values and you will get 85, which has prime factors of 17 and 5. Adding 17, 5, and 2 you then arrive at 24.
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Re: If a, b, and c are positive consecutive odd integers, what is the [#permalink] New post   19 Oct 2022, 09:33
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Barfi wrote:
Can someone please explain to me why we take the prime factor for 2^a i.e. 2^123 as 2?


What else could it be? Exponentiation (when both the base and the exponent are positive integers) does not "produce" primes. So for example, the primes of 30^n, where n is a positive integer, are the same as the primes of 30, which are 2, 3, and 5.

2^123 = 2*2*...*2, what other prime could magically appear there?

Gmat Club Official Explanation:

If a, b, c and d are consecutive odd integers such that \(123 < a < b < c < d\), then what is the sum of the distinct prime factors of \(2^a + 2^b + 2^c + 2^d\) ?

A. 17
B. 19
C. 22
D. 24
E. 25

Since a > 123, then a is a positive integer.

Next, since a, b, c and d are consecutive odd integers and \(a < b < c < d\), then:

\(2^a + 2^b + 2^c + 2^d = 2^a + 2^{a + 2} + 2^{a + 4} + 2^{a + 6} = 2^a(1 + 2^2 + 2^4 + 2^6) = 2^2*85=2^a*5*17\).

The sum of the distinct prime factors of \(2^a*5*17\) is 2 + 5 + 17 = 24.

Answer: D.
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Re: If a, b, and c are positive consecutive odd integers, what is the [#permalink] New post   19 Oct 2022, 09:36
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Round2Hopeful wrote:
Answer should be 24. The odd consecutive factors are 123, 125, 127 and 129. Plug them into the equation. You have 2^123 + 2^125 + 2^127+2^129. Factor out 2^123 and you will then have (1+2^2+2^4+2^6) in the equation. Sum those values and you will get 85, which has prime factors of 17 and 5. Adding 17, 5, and 2 you then arrive at 24.


Two notes:

1. 123 < a < b < c < d, so a cannot be 123.

2. 123 < a < b < c < d does not necessarily means that a is the next odd number after 123. Yes, a, b, c, and d can be 125, 127, 129, and 131 but they can also be any other four consecutive odd numbers greater than 123, for example, 22137, 22139, 2241 and 22143.

Hope it helps.
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Re: If a, b, and c are positive consecutive odd integers, what is the [#permalink] New post   29 Oct 2023, 13:19
Substitute some numbers since this is a manageable summation.

1,3,5,7

2+8+32+128=170=2*5*17

2+5+17=24

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Re: If a, b, and c are positive consecutive odd integers, what is the [#permalink]
29 Oct 2023, 13:19
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