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Bunuel
What is the units digit of positive integer x?

(1) 7, 15, 24, and n are all factors of x.
(2) 6! < x < 7!

B is out straight away

A tells that number can end in 0/5

C is the answer i believe'

HI,
Always be careful while eliminating a statement..
what is statement I
(1) 7, 15, 24, and n are all factors of x.
the factors are 15 and 24, so x will be a MULTIPLE of 5 and 2..
thus UNIT's digit will be 0
Suff
ans A


Ya agreed sir...will be careful.
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Since x is divisible by both 15 and 24 from statement 1 we can say that unit digit of x must be zero.
Again statement 2 is not sufficient as we don't know what x might be.
hence A
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Great Question.
Here we need the units digit of x
Statement 1=>
this statement tells us that 5 and 2 must be prime factors of x.
Hence the units digit of x must be zero.
Hence Sufficient.

Statement 2
x must lie between (720,720*7)
Not sufficient

Hence A
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What is the units digit of positive integer x?

1) 7, 15, 24, and n are all factors of x

2) 6! < x < 7!
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1) with 15,24 ... the unit digit has to be 0
Sufficient
1) from Too wide a range ... not sufficient

A

Posted from my mobile device
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1) 7, 15, 24, and n are all factors of x
Analysis:
x is multiple of 7, 15, 24, and n

Simplify:
x = 7*15*24*n
Unit digit of x = Product of unit digits of the given numbers = 7*5*4*n, which will be 0 as 5*4 is 20

Sufficient.

2) 6! < x < 7!
Analysis:
720<x<5040
Unit digit of x is unknown.
Insufficient

A is correct.

Concept:
Unit digit of product of given numbers = Product of unit digits of the given numbers
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What is the units digit of positive integer x?

1) 7, 15, 24, and n are all factors of x

2) 6! < x < 7!

Merging topics. Please check the discussion above. Hope it helps.
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S1: Taking LCM(7,15,24,n) gives us that x is a multiple of 840n. Hence units digit is always 0. Sufficient.

S2: x lies between 6! and 7! - means multiple values. Insufficient.

Hence, option A.
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