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b, c, d, e, f and g are different positive integers. When b is divided

BrentGMATPrepNow

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b, c, d, e, f and g are different positive integers. When b is divided by c, the remainder is d. When e is divided by f, the remainder is g. What is the value of d+g?

(1) b + e = 16
(2) c + f = 6

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chetan2u

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b, c, d, e, f and g are different positive integers. When b is divided by c, the remainder is d. When e is divided by f, the remainder is g. What is the value of d+g?

When b is divided by c, the remainder is d. ----- $b=cr+d$
When e is divided by f, the remainder is g. ----- $e=fs+g$

(1) b + e = 16
Various combination possible...

(2) c + f = 6
Now b>c>d and e>f>g
C or f cannot be 1, then the remainder will be 0..
various possibilities of c and f when c+f=6..
a) c=2 and f=4... d and g will be 1 and 3 in any order but d+g=1+3=4
b) c=4 and f=2... d and g will be 1 and 3 in any order but d+g=1+3=4
No other possibility as c=f=3 is not possible as all are supposed to be different positive integers
d+g=4
suff

B

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18 Sep 2019, 18:29

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Mo2men

chetan2u

Dear chetan2u

I do not understand why g could be 3. If f=4, then reminder could be any number from 1, 2 or 3 (all are less than the quotient). Only possible value of d is 1.
So we have various possibility that d+g= 2, 3 or 4

Where did I go wrong?

Hi,

The reason is that all numbers are DIFFERENT positive integers.
So, if c+f=2+4...... the remainders d and e cannot be 2 and 4..

If f=4, then c=2.. The remainder in case of c will be 1 as it has to be less than c..
Now, when f=4, remainder cannot be 1 and 2, since these two values have already been taken

General Discussion

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17 Sep 2019, 07:54

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d+g =4

Since possible combinations for c+f =6 will give sufficient cases such that d+g will always be 4 in the above three cases

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17 Sep 2019, 09:46

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GMATPrepNow

b, c, d, e, f and g are different positive integers. When b is divided by c, the remainder is d. When e is divided by f, the remainder is g. What is the value of d+g?

(1) b + e = 16
(2) c + f = 6

given b=a*r+d
and e=x*f+g
#1
b + e = 16
there can be many possible values where b+e=16 so not sufficient
#2
c + f = 6
we know b>c>d and e>f>g
also that all of them are different integers
so only possiblities are ; 5>2>1 and 7>4>3 there are no other possibilities ; except value pf c & f being interchanged
so d+g = 4 sufficeint
IMO B

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17 Sep 2019, 17:52

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Given:

Factor of 1: B = C + D & E= F +G <=> b + e = (f+c) + (d+g)
Factor of 2: B = 2C + D & E = 2F + G <=> b + e = 2(f+c) + (d+g)
Factor of 1 & 2: B = 2C + d & E = F + G <=> b + e = (2c + f) + (d+g)

1. b+e = 16
INS because we don't know anything about c and f

2. c+f= 16
INS similarly, we don't know b & e

1+2:

Scenario 1: 16 = 6 + d + g (ok)
Scenario 2: 16 = 12 + d + g (ok)
Scenario 3: 16 = 2f+c + (d+g) (undetermined)

=> E. Neither

Correct me if I wrong please.

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Updated on: 18 Sep 2019, 19:32

Originally posted by Kinshook on 18 Sep 2019, 19:17.
Last edited by Kinshook on 18 Sep 2019, 19:32, edited 2 times in total.

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GMATPrepNow

b, c, d, e, f and g are different positive integers. When b is divided by c, the remainder is d. When e is divided by f, the remainder is g. What is the value of d+g?

(1) b + e = 16
(2) c + f = 6

b, c, d, e, f and g are different positive integers. When b is divided by c, the remainder is d. When e is divided by f, the remainder is g. What is the value of d+g?
Since b,c,d,e,f are different positive integers
b = kc + d; b>c>d
e = lf + g; e>f>g

(1) b + e = 16
Many dIfferent combinations are possible.
NOT SUFFICIENT

(2) c + f = 6
c+f = 6;
(c,f) = {(2,4),(4,2)} since d, g are positive integers. & c>d; f>g; Combination (3,3) is not possible since c & f are different.
(d, g) = {(1,3),(3,1)}
d + g = 1+3 = 4
SUFFICIENT

IMO B

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18 Sep 2019, 19:27

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chetan2u

how can we say b>c>d ? b can be 3 and c can be 4 for b/c remainder can be 3

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Peddi

chetan2u

how can we say b>c>d ? b can be 3 and c can be 4 for b/c remainder can be 3

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Peddi
When 3 is divided by 4, remainder is 3, b=3=d in this case which is invalid.
But it is given in the question that
b, c, d, e, f and g are different positive integers.
If b<c then d=b which is not valid
b>c>d is valid

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19 Sep 2019, 15:03

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GMATPrepNow

b, c, d, e, f and g are different positive integers. When b is divided by c, the remainder is d. When e is divided by f, the remainder is g. What is the value of d+g?

(1) b + e = 16
(2) c + f = 6

Given: b, c, d, e, f and g are different positive integers. When b is divided by c, the remainder is d. When e is divided by f, the remainder is g.

Target question: What is the value of d+g?

Statement 1: b + e = 16
There are several cases that satisfy statement 1. Here are two:
Case a: b = 5, c = 2, d = 1, e = 11, f = 4 and g = 3. In this case, the answer to the target question is d + g = 1 + 3 = 4
Case b: b = 5, c = 2, d = 1, e = 11, f = 7 and g = 4. In this case, the answer to the target question is d + g = 1 + 4 = 5
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: c + f = 6
If c + f = 6, then it MUST be the case that one value (c or f) is 4 and the other value is 2.
How do we know this?
Well, c and f can't both equal 3, since all 6 values are DIFFERENT
Also, neither value can be 0, since we're told all 6 values are POSITIVE
And, the numbers cannot be 1 and 5, because if we divide by 1, we get a remainder of 0, and we're told all 6 values are POSITIVE.
So, the two values must be 2 and 4.

---ASIDE---
Useful remainder property
When positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < D
For example, if we divide some positive integer by 7, the remainder will be 6, 5, 4, 3, 2, 1, or 0
-----------

When we apply the above property, we can conclude that d < c and g < f
So, the GREATEST possible value of d+g is 4.
We also know that d+g cannot add to 3, because the only way to get a sum of 3 is for the numbers to be 1 and 2, and we already
accounted for the 2 when we concluded that one one value (c or f) is 4 and the other value is 2.
Using similar logic, we can show that d+g cannot add to 2 or 1 either.
So, it MUST be the case that d + g = 4
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Aside: here's one way to for d+g to equal 4:
b = 5, c = 2, d = 1, e = 11, f = 4 and g = 3. In this case, the answer to the target question is d + g = 1 + 3 = 4

Cheers,
Brent

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09 Oct 2019, 05:56

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GMATPrepNow

b, c, d, e, f and g are different positive integers. When b is divided by c, the remainder is d. When e is divided by f, the remainder is g. What is the value of d+g?

(1) b + e = 16
(2) c + f = 6

given: all different positive integers;
rem: b/c is d… 0<d<c; also, b>c since b<c would give a remainder equal to b, or b=d.
rem: e/f is g… 0<g<f; also, e>f since e<f would give a remainder equal to f, or e=f.
so: 0<d<c<b and 0<g<f<e

(1) b + e = 16: insufic.
b=9, c=8, d=1; e=7, f=5, g=2; d+g=3;
b=9, c=5, d=4; e=7, f=3, g=1; d+g=5;

(2) c + f = 6: sufic.
(c,f)=(2,4); then 0<d<c<b and 0<g<f<e = 0<d<2 and 0<g<4… and d has to be d=1, so g={≠1,≠2}=3; d+g=4.
(c,f)=(4,2); then 0<d<c<b and 0<g<f<e = 0<d<4 and 0<g<2… and g has to be g=1, so d={≠1,≠2}=3; d+g=4.

Answer (B)

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03 Feb 2020, 22:37

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Statement 1
b+e =16. From this info, various combinations will come up with 16 (i.e 2+15, 3+13, 4+12 etc as long as the sum is 16 and 1 is not a possible value).

Statement 2
c+f=6
6= 2+4 or 4+2. It is not possible to be 1+5 or 5+1 as c & f are denominators that gives a remainder.

If c=2, f=4
d=1. g which is a remainder after division of 4 can be 1,2 or 3. However, since it is mentioned that b, c, d, e, f and g are different positive integers, the remainder can only be 3 since 1 & 2 is taken by c & d respectively.

If c=4, f=2
g=1. d which is a remainder after division of 4 can be 1,2 or 3. However, since it is mentioned that b, c, d, e, f and g are different positive integers, the remainder can only be 3 since 1 & 2 is taken by g & f respectively.

At either option, it will be 4+2=6

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chetan2u

why 0 is not possible as a remainder ?

chetan2u

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Rohan271

chetan2u

why 0 is not possible as a remainder ?

All variables are positive integers, so none of them will be 0, as 0 is neither positive nor negative.
So remainder will not be 0.

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