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tjkim88
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tjkim88
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tjkim88
If n is the product of three consecutive integers, and n is divisible by 7, which of the following is not necessarily a divisor of n ?


(A) 6

(B) 14

(C) 21

(D) 28

(E) 42

can you explain why???

Hi,
first you have to see the minimum value and ways..
1) 5*6*7= 2*3*5*7
2) 6*7*8= 2^4*3*7
3) 7*8*9..=2^3*7*9..

what is the common in the three terms..
2*3*7..

lets find the choice which contains any term beyond these three..
(A) 6
2*3= so Always

(B) 14
2*7= again always

(C) 21
3*7= again always

(D) 28
2*2*7..
may be there, may not be there
CORRECT

(E) 42
2*3*7= again always

ans D
hope it helps

thx a lot I completely understand it !
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tjkim88
If n is the product of three consecutive integers, and n is divisible by 7, which of the following is not necessarily a divisor of n ?


(A) 6

(B) 14

(C) 21

(D) 28

(E) 42

can you explain why???

28 divides only if the sequence includes a multiple of 4
it works for 6,7,8 and 7,8,9, but not for 5,6,7
D
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n = (x)(x+1)(x+2)

if we know that this is divisible by 7, then we need to rationalize the other prime factors

at a minimum, we have (1)(2)(3) which means we have at least one 2 and one 3 as prime factors FOR SURE.
7*1 = 7 given
7*2 = 14
7*3 = 21
7*2*3 = 42

We are not guaranteed a 2nd 2 which is what is needed for 28 -->7*2*2 = 28
So 28 is the answer.
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