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If a and b are non-negative integers, is the number of the divisors of

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If a and b are non-negative integers, is the number of the divisors of  [#permalink]

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New post 04 Nov 2019, 03:29
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A
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C
D
E

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Question Stats:

26% (01:18) correct 74% (01:18) wrong based on 34 sessions

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Re: If a and b are non-negative integers, is the number of the divisors of  [#permalink]

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New post 04 Nov 2019, 04:02
a, b —non-negative integers (includes zero)

# of divisors of a > # of divisors of b ???

(Statement1) a > b
—> 8> 3 (yes)
—> 6 >5 (No)
Insufficient

(Statement2): 2b= a
2*3 = 6 ( 2< 4 —yes)
2*10= 20 (4 < 6– yes)
But
If a= 0, b=0, then the number of divisors of a and b will be equal to each other—NO)

Insufficient

Taken together 1&2,
a >b , 2b= a
—> both a and b cannot be zero. (a>0, b>0)

Well, if b =3, then a = 6, (yes)
If b= 25, then a= 50, (yes)

Sufficient.

The answer is C

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If a and b are non-negative integers, is the number of the divisors of  [#permalink]

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New post 04 Nov 2019, 05:21
Question is asking if a has more factors than b has. I would be inclined to take only positive divisors; so Im flagging it out early that I may be wrong, but since the question did not state specifically for the divisors to be only positive, I will include negative divisors as well.

(1) if a = non-prime eg 6, b = 5, a has 8 divisors (1,2,3,6,-1,-2,-3,-6) while b has 4 (-5,-1,1,5), answer is yes.

But if a = 7, while b = 6, then the answer is no.

(2) If 2b = a, b is a factor of a, and if b > 0, then a would always have more factors than b.

However, if b = 0 i.e still a non-negative integer, then a = 2*0 = 0 where a does not have more factors than b.

(1+2) Since a > b, and 2b = a, then b cannot = 0 since 2*0 = a = 0 (which does not comply with (1) here).

Therefore b > 0 ; if b is positive eg 5, then a = 10 (a has more factors) - Answer is YES.

(C)

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If a and b are non-negative integers, is the number of the divisors of   [#permalink] 04 Nov 2019, 05:21
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If a and b are non-negative integers, is the number of the divisors of

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