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

Question

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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Bunuel · Accepted Answer

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.

Round2Hopeful · Answer

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.

bellbell · Answer

*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 .

Bunuel · Answer

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.

Regor60 · Answer

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

Posted from my mobile device

bumpbot · Answer

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

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

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