If integer n is divisible by 25, is the tens digit of n less

Question

If integer n is divisible by 25, is the tens digit of n less than 5?

(1) The square root of n is an integer whose units digit is 0.
(2) 4n has 40% more factors than does n.

ArvGMAT · Answer

a is sufficient. 
if the units digit is 0, and number is square of an integer, the number is 
100,400,900 etc. so, the tens digits is also 0<5

b is not sufficient.
Take 75 for example - it has 2 factors 5 and 3. 
4n is 300 and has 5,3,and 2 as factors - 50% more factors 
tens digit is 7>5

now 225 also has 2 factors 5 and 3
4n has 3 factors 50% more
tens digit is 2<5

so, A is sufficient.
Anyone else care to verify???

gmatornot · Answer

Take 75 for example - it has 2 factors 5 and 3 :no

it has only 2 prime factors... I think the answer is B or E From ST 1 ^2 = integer with units digit 0, possibilities 50,100, 150, 200 etc... As we can see the the tens digit could be 0 or 5 hence NOT SUFF. Statement 2 gives a long list of factors.... :cry:

Professor · Answer

good question i have little confusion.

from (1), if sqrt of n has an integer with unit digit 0 means n is a multiple of 100 whose tens digit is always <5. however n could be 0 whose tens digit, i suppose, is 0<5.

from (2), n could be 100 or 625 both have 40% more factors if they are multiplied by 4.

if n = 100 = 1x2x2x5x5. altogather 5 factors
4n = 4(100) = 1x2x2x2x2x5x5. altogather 7 factors and increased by 40% (2/5).

if n = 150 = 1x2x3x5x5. altogather 5 factors
4n = 4(150) = 1x2x2x2x3x5x5. altogather 7 factors and increased by 40% (2/5).

if n = 100, its tens digit is <5. if n is 150, its tens digit is equal to 5. so it is clear A.

gmatornot · Answer

I second Professor's answer.. A seems like the correct choice here (after the correction ) :-D

heman · Answer

(1) n= 100 Units digit<5 Yes
 n = 1000 Units digit<5 Yes
 AD

(2) n= 5^2 a where a is an int 
 4n = 2^2 5^2 ??

Guessing A

Heman

Professor · Answer

hmmmmmmm.....................
actually i was facorizing n.

i am sure A is it and B doesnot work but i am not sure whether B is a valid statement or not. since n is divisibly by 25, 4n cannot have only 40% more factors than n has.

need clearification. :?:

prashrash · Answer

statement B states "4n has 40% more factors than n". It doesn't say if 4n can have more than 40% factors or not. As old_dream_1976 pointed out, 400 (4x100) has 11 factors as compared to 100 which has 7 factors (1,2,4,10,20,25,50).
This is an increase of 4/7% which is approximately 57%. Hence we cannot determine the soln. to the problem based on stmt (2). Hence answer should be A.

Thanks old_dream_1976 for pointing me in the right direction (I hope it is the right direction

). Haven't quoted you in the interest of keeping this post short and sweet.

kevincan · Answer

How do we get the number of factors? 100 has 9 factors and 400 has 15 factors .

2times · Answer

Ans is D 

Statement A has already been shown to give the ans.... refer to 'ArvGMAT' Above

now for statement (2)

4n gives 40% more factors than n.
=> 2^2*n gives 40% more factors.

let n = 2^a*(N') 
then 4n = 2^(a+2)*(N')

now let us assume the # of factors of N are Z; # of factors of 4N are 1.4Z
 ]

so lets say the # of factors of n are (a+1)*(b+1)....
# of factors of 4n are (a+3)*(b+1) ....

also (a+3)/(a+1) = 1.4/1
=> a= 4

=> in the no n 2 ^ 4 ; so n is basically 16N' 
now we know that N is divisible by 25; that means that the power of 5 is even

so we can easily write n = 2^4*5^b* N'' ; where b is an even no

we also know that if 5 raised to a power > 1; last two digits will always be 25, so when 5^b; last 2 digits = 25.

so if you see the no it is of a form where the powers of 2 and 5 give last 2 digits as zero... 
example let b = 2
so the number n = 2^4*5^2*N'' (where N'' are the remaining factors of N)
so this no will take the form 25*16 = 400*N'' => last two digits are zero; now for all values of N'' & b(such that it is an even positive number) last two digits will always be zero...

therefore  D 

Kevincan OA??

Vel · Answer

I'm not sure about "A". how about 50 - for example

prashrash · Answer

2times, could you please explain how you arrived at 1.4Z and also (a+3) as the first factor of 4n ?

2times · Answer

1.4 is 140% of 1
# of factors of 6n are 40% greater than # of factors of n

(a+3) is not a factor; it is (1+ the power to which 2 is raised to) and it leads to the total no of factors in 6n!

If integer n is divisible by 25, is the tens digit of n less

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