HOT Competition 2 Sep/8PM: If k is a positive integer and 175 divided

Question

If k is a positive integer and 175 divided by k leaves a remainder of 7, what is the value of k?

(1) The product of any two factors of k is odd
(2) The sum of any two factors of k is even.

M36-104

Bunuel · Accepted Answer

Official Solution:

If  k  is a positive integer and 175 divided by  k  leaves a remainder of 7, what is the value of  k ? 
 
175 divided by  k  leaves a remainder of 7:  175=kq + 7  ( notice here that  k  (the divisor) must be greater than 7 (the remainder)) : 
 
 168=kq ;
 
 kq=2^3*3*7 .
 
(1) The product of any two factors of  k  is odd.
 
 k  cannot be even because if it is then it has 1 and 2 as factors and  1*2=2=even . Thus  k  must be odd. In this case all its factors will be odd and the product of any two distinct positive factors of  k  will be odd. Now, if  k=odd  and  k > 7 , then from  kq=2^3*3*7  it follows that  k=21  (the only odd factor of 168 greater than 7). Sufficient.
 
(2) The sum of any two factors of  k  is even.
 
 k  cannot be even because if it is then it has 1 and 2 as factors and  1*2=2=even . Thus  k  must be odd. In this case all its factors will be odd and the sum of any two distinct positive factors of  k  will be even. Now, if  k=odd  and  k > 7 , then from  kq=2^3*3*7  it follows that  k=21  (the only odd factor of 168 greater than 7). Sufficient.

Answer: D

Kinshook · Answer

Asked: If k is a positive integer and 175 divided by k leaves a remainder of 7, what is the value of k?

175 =km + 7; where m is an integer
168 = km = 2^3×3×7
Since division by k, leaves remainder 7; k>7

(1) The product of any two factors of k is odd
k is odd; k>7 and km=168
k =21 is the only solution since k is odd, k>7 and I is a factor of 168.
SUFFICIENT

(2) The sum of any two factors of k is even.
k is odd since sum of 2 odd numbers is even, k cannot be even since 1 is a factor of k and is odd.
k is odd; k>7 and km=168
k =21 is the only solution since k is odd, k>7 and I is a factor of 168.
SUFFICIENT

IMO D

Posted from my mobile device

bidskamikaze · Answer

If k is a positive integer and 175 divided by k leaves a remainder of 7, what is the value of k?

we have 175 = k*x + 7
k*x = 168
 168 = 2^3 * 3 * 7 
possible factors: 21 and 8 (other factors 1,2,3,4, 7 are <= remainder 7, hence not possible)

(1) The product of any two factors of k is odd
- The leaves only 21 as the possible answer; product of factors of 8 is always even
SUFFICIENT

(2) The sum of any two factors of k is even.
- This also leaves only 21 as possible answer, because the condition is not true for 8 + 1 = 9
SUFFICIENT

Answer D

Varunsawhney8 · Answer

If k is a positive integer and 175 divided by k leaves a remainder of 7, what is the value of k?
From Given Data
We Know
175= KX+7

So From Here we know K should be Greater than 7

also
we know
175-7=KY
So
KY=168

Possible Values

84 X 1
56 X 3
42 X 4
28 X 6
24 X 7
21 X 8

So K can Have all values greater than 7

This means

K can be 8,21,24,28,42,56,84

(1) The product of any two factors of k is odd

If Product of any two Integer is odd K should be Odd integer itself

Hence K=21

Statement 1 alone is sufficient

(2) The sum of any two factors of k is even.

This means all the factors should be odd
It will only be possible when K is an odd integer
as if K is even integer
For Eg:
if K is 24 which has factors - 1,2,3,4,6,8,12,24
Then 1+24 will be odd
But When K is 21 which has factors- 1,3,7,21

Then by selecting any 2 values we will have odd sum

Hence K cannot be even integer and K= 21
Statement 2 is sufficient

Answer: D

Posted from my mobile device

Nevernevergiveup · Answer

If k is a positive integer and 175 divided by k leaves a remainder of 7, what is the value of k?

175=k( )+7  &  k>7 
 k( )=168  so we can infer that k is a multiple of 168(factor form  168=3x7x(2^3) )

(1) The product of any two factors of k is odd

k = 21(3x7)  as  k>7  and cant have even factors(any two)

SUFFICIENT

(2) The sum of any two factors of k is even.

sum of two factors is even so either both even or odd

only possible value is 21(its factors are off so sum is even)

also 8 can be considered but factors 2+4=even and 1+4 is odd

SUFFICIENT

SO D is answer

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

HOT Competition 2 Sep/8PM: If k is a positive integer and 175 divided

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

HOT Competition 2 Sep/8PM: If k is a positive integer and 175 divided

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards