When 85 is divided by a positive integer n, the remainder is 7, and wh

Question

When 85 is divided by a positive integer n, the remainder is 7, and when 85 divided by 2n, the remainder is n + 7. How many different values can n take ?

A. 0
B. 1
C. 2
D. 3
E. 4

av1901 · Accepted Answer

Surely! Ok so we are given that 85 when divided by  n  gives us a remainder of 7 and when divided by  2n  gives us a remainder of  n+7

85 = n*C + 7  and  85 = (2n)C + (n+7)  {Remainder expressions where C represents the Quotient}

Now  85  gives a remainder of  7  when divided by  n  => 2 things

a)  n > 7  (only then can it have a remainder greater than 7)
b)  n  is a factor of  78  (because only a number greater than 7 which is a factor of 78 can leave a remainder of 7 when it divides 85)

Now we know that  n  is a factor of  78 , so we find all factors of  78

78 = 2*3*13 , Factors =  1, 2, 3, 6, 13, 26, 39, 78

n > 7 so 1, 2, 3, 6  are out. Let us examine the other options

If  n = 13: 2n = 26 : Remainder ( \frac{85}{2n} ) =  16 ≠ n+7 (13+7=20) 
If  n = 26: 2n = 52 : Remainder ( \frac{85}{2n} ) =  33 = n+7 (26+7=33) 
If  n = 39: 2n = 78 : Remainder ( \frac{85}{2n} ) =  7 ≠ n+7 (39+7=46) 
If  n = 78: 2n = 156 : Remainder ( \frac{85}{2n} ) =  85 = n+7 (78+7=85)

2 possible values satisfy. Hence   Answer = 2 (C)

Hope its clear

Bunuel · Answer

When 85 is divided by a positive integer  n , the remainder is 7, and when 85 is divided by  2n , the remainder is  n + 7 . How many different values can  n  take?

A. 0
B. 1
C. 2
D. 3
E. 4

Dividing 85 by  n , and obtaining a remainder of 7, can be expressed as  85 = nq + 7 . 
 
Similarly, dividing 85 by  2n , and obtaining a remainder of  n+7 , can be expressed as  85 = 2np + (n+7) .
 
The first equation,  85 = nq + 7 , simplifies to  nq = 78 , indicating that  n  is a factor of 78. The factors of 78 are 1, 2, 3, 6, 13, 26, 39, and 78. However, since  n  must be greater than 7 (because the divisor is always larger than the remainder), the possible values for  n  are 13, 26, 39, or 78.
 
The second equation,  85 = 2np + (n+7) , implies that  n+7  must be odd because  2np  is even and their sum must equal the odd number 85. Thus,  n  must be even. This narrows the possible values of  n  to 26 and 78.
 
For  n=26 , dividing 85 by  2*26 = 52  leaves a remainder of 33, which is equal to  n+7=26+7 . Hence, this value meets the conditions.
 
For  n=78 , dividing 85 by  2*78 = 156  leaves a remainder of 85, which is equal to  n+7=78+7 . Hence, this value meets the conditions.
 
If  n=78 , then 85 divided by  2*78=156  gives the reminder of 85, which does equal to  n+7=78+7 . So, this value satisfies the conditions.
 
Therefore,  n  can take two values: 26 and 78.

Answer: C

samarth1222 · Answer

hi 
can someone provide the solution to this question I was not able to solve it ?

fsf4 · Answer

Hi
Same here, I got B

Bunuel · Answer

av1901 · Answer

Yes Bunuel, thanks for pointing that out, should have assigned a different variable for the other quotient. Will keep in mind from now.
Also question was fun, thanks for these links to fresh Gmat Club Tests questions

Posted from my mobile device

sachi-in · Answer

When 85 is divided by a positive integer n, the remainder is 7, and when 85 divided by 2n, the remainder is n + 7. How many different values can n take ?

A. 0
B. 1
C. 2
D. 3
E. 4

Multi-part reasoning:

85 = n*a + 7 = 2n*b + n + 7

n*a = n*(2b + 1) = 85 - 7 = 78 = 2 * 3 * 13

We need to break 78 into pairs that satisfy n * (2b + 1)

78 * 1
39 * 3
26 * 2 --> not possible 2b + 1 has to be odd
13 * 6 --> not possible 2b + 1 has to be odd

And we are not going below 13 since we know n > 7 as it leaves 7 remainder so putting 6, 3, 2, 1 in it's place is unnecessary.

Only two pairs of n and b satisfy this equation.

Vordhosbn · Answer

From the first statement we know n has to be bigger than 7 (1).
From the second statement we get:

quotient times divisor + remainder is 85.

q * (2n) + (n + 7) = 85 -> 
n (2q +1 ) = 78 
This simply means that n is a multiple of 78, and the multiplier of it is odd, because (2q + 1) is always odd.

78 = 13 * 3 * 2 * 1

Thus (2q + 1) can be either 1, 3 or 13:

if 2q + 1 is 1 then n is 78 which satisfies (1) - one possible answer
if 2q + 1 is 3 then n is 26 which again satisfies (1) - another possible answer
if 2q+1 is 13 then n has to be 6 which cannot be the case because of the first statement (1)

Thus, 2 possible solutions for n: 78 or 26

Kinshook · Answer

When 85 is divided by a positive integer n, the remainder is 7, and when 85 divided by 2n, the remainder is n + 7.

How many different values can n take ?

85 = nk + 7; nk = 85-7 = 78; where k is an integer
78 = 2*3*13

85 = 2nk' + n + 7 = nk + 7 
k = 2k' + 1 = 78/n

k = {3, 13}
n = {26, 6}

IMO C

juliansorre · Answer

85 = kn + 7 (1)
85 = 2pn + n+ 7 (2)

Therefore, 
2pn + n + 7 = kn + 7
n(2p+1-k) = 0 (3)

From (1), we know that n > 7.

Thus,
k= 2p + 1, which is odd

if k is odd then n must be even

from (1), we get
kn = 78

factoring 78 = 2*3*13
Knowing that n>7 and n is even

n can assume only two values: 2*13 or 2*3*13

C

Kinshook · Answer

When 85 is divided by a positive integer n, the remainder is 7, and when 85 divided by 2n, the remainder is n + 7.

How many different values can n take ?

85 = nk + 7 = 2nk' + n+7
2nk' + n = nk
k = 2k'+1; k is odd

78 = nk = 1*2*3*13 = (2k'+1)n; k is odd; n is even
Since when 85 is divided by a positive integer n, the remainder is 7, n>7

n = {26,78} 2 values

IMO C

When 85 is divided by a positive integer n, the remainder is 7, and wh

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