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b, c, d, e, f and g are different positive integers. When b is divided
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17 Sep 2019, 06:36
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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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Re: b, c, d, e, f and g are different positive integers. When b is divided
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17 Sep 2019, 07:45
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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b, c, d, e, f and g are different positive integers. When b is divided
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17 Sep 2019, 07:54
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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Re: b, c, d, e, f and g are different positive integers. When b is divided
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17 Sep 2019, 09:46
GMATPrepNow wrote: 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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Re: b, c, d, e, f and g are different positive integers. When b is divided
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17 Sep 2019, 17:52
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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Re: b, c, d, e, f and g are different positive integers. When b is divided
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18 Sep 2019, 17:05
chetan2u wrote: 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 Dear chetan2uI 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?



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Re: b, c, d, e, f and g are different positive integers. When b is divided
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18 Sep 2019, 18:29
Mo2men wrote: chetan2u wrote: 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 Dear chetan2uI 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
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b, c, d, e, f and g are different positive integers. When b is divided
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Updated on: 18 Sep 2019, 19:32
GMATPrepNow wrote: 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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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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Re: b, c, d, e, f and g are different positive integers. When b is divided
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18 Sep 2019, 19:27
chetan2u wrote: 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 how can we say b>c>d ? b can be 3 and c can be 4 for b/c remainder can be 3 Posted from my mobile device



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b, c, d, e, f and g are different positive integers. When b is divided
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18 Sep 2019, 22:49
Peddi wrote: chetan2u wrote: 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 how can we say b>c>d ? b can be 3 and c can be 4 for b/c remainder can be 3 Posted from my mobile device PeddiWhen 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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Re: b, c, d, e, f and g are different positive integers. When b is divided
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19 Sep 2019, 15:03
GMATPrepNow wrote: 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 = 4Case 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 = 5Since 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 propertyWhen positive integer N is divided by positive integer D, the remainder R is such that 0 ≤ R < DFor 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 = 4Cheers, Brent
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b, c, d, e, f and g are different positive integers. When b is divided
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09 Oct 2019, 05:56
GMATPrepNow wrote: 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)




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