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Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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08 Sep 2015, 23:45
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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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09 Sep 2015, 00:23
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Bunuel wrote: Each bead in an urn is marked with a distinct positive integer and colored according to that integer’s remainder after division by 5, as shown in the following table:
Remainder Color 0  Red 1  Blue 2  Green 3  Yellow 4  Orange
Four blue beads, three green beads, two yellow beads, and one orange bead are withdrawn. If the product of the numbers on these beads is displayed on another bead, according to the rules above, then the color of that bead is…
(A) red (B) blue (C) green (D) yellow (E) orange
Kudos for a correct solution. Solution : Red beads will be of the form 5k, blue  5k+1 , green  5k+2 , yellow  5k+3, orange  5k+4 . So, product will be (5k1+1)(5k2+1)(5k3+1)(5k4+1)(5k5+2)(5k6+2)(5k7+2)(5k8+3)(5k9+3)(5k10+4) Which will effectively be 5(p) + (1)(1)(1)(1)(2)(2)(2)(3)(3)(4) = 5(p) + (1)(8)(9)(4) = 5(p) + 8 = 5(p) + 3. Therefore, Yellow. Otherwise, just put the values 5 for red, blue  1 , green  2 , yellow  3, orange  4. Product = (1)(1)(1)(1)(2)(2)(2)(3)(3)(4) = 288 = remainder 3. Yellow. Option D.



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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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10 Sep 2015, 05:49
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 Bluebead = 5K +1, so 4*BB = 5*(K*4) + 4 = 5*X + 4 (X=K*4) Greenbead = 5*V +2, 3*GB= 5*(3*V) + 6 = 5*(3*V +1) + 1 = 5*Y + 1 Yellow Bead = 5*M+ 3, 2 YB= 5*(2*M) + 6 = 5*(2*M + 1) + 1 = 5*W + 1;  1 Orange bead = 5*Z + 4; By multiplying all the terms together, we get as a result a remainder of 1,hence the correct answer is (B).



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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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11 Sep 2015, 20:38
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Each bead in an urn is marked with a distinct positive integer and colored according to that integer’s remainder after division by 5, as shown in the following table: Remainder Color 0  Red 1  Blue 2  Green 3  Yellow 4  Orange Four blue beads, three green beads, two yellow beads, and one orange bead are withdrawn. If the product of the numbers on these beads is displayed on another bead, according to the rules above, then the color of that bead is… (A) red (B) blue (C) green (D) yellow (E) orange It is a remainder problem : There is simple logic behind the remainder of number when divided by another integer and again using that remainder to get the final . Because from remainder you guess the number ... and thus taking the remainder or number  will result in the same ans . ( may be my explanation sounds bit odd  Bunuel hope you got what i am trying to say ... please explain it if possible ) As per the question i will only use the remainder to get the answer to this question . Four blue beads, three green beads, two yellow beads, and one orange bead are withdrawn. 4 blue = 1*1*1*1 3 green =2*2*2 2 yellow =3*3 1 orange =4 On multiplying the above and dividing by 5 we get our remainder as 3 . Which is yellow . Hence ans D
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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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13 Sep 2015, 07:58
Bunuel wrote: Each bead in an urn is marked with a distinct positive integer and colored according to that integer’s remainder after division by 5, as shown in the following table:
Remainder Color 0  Red 1  Blue 2  Green 3  Yellow 4  Orange
Four blue beads, three green beads, two yellow beads, and one orange bead are withdrawn. If the product of the numbers on these beads is displayed on another bead, according to the rules above, then the color of that bead is…
(A) red (B) blue (C) green (D) yellow (E) orange
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:The key to this problem is to translate the “game” into more familiar mathematical language. First of all, each bead’s color corresponds to a remainder after division by 5. For instance, Red = R0 (remainder of 0), which means that the number in question is a multiple of 5. Blue = R1, Green = R2, etc. We withdraw four blues, three greens, two yellows, and an orange. In other words, we have four R1 numbers, three R2 numbers, two R3 numbers, and one R4 number. If we want to, we can pick actual numbers. It would be best to pick small numbers—for instance, R1 could actually be 1, because when you divide 1 by 5, you get a quotient of 0 and a remainder of 1. Likewise, R2 could be 2, R3 could be 3, and R4 could be 4. Multiplying these numbers together, we get 1x1x1x1x2x2x2x3x3x4 = 8x9x4 = 72×4 = 288. The remainder after division by 5 would be 3, and the color of the bead would be yellow. We would get the same result without picking numbers, of course—we would have to multiply the remainders together, which would give us R288, and then we’d reduce that to R3. Either way, we have a yellow bead. The correct answer is D.
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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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21 Mar 2016, 14:30
I think this question should be more precise and mention that the numbers that are marked are actualy the reminders. To me was not clear what the author wants to say. does anyone else had this issue?



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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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23 Mar 2016, 11:04
SIMPLY 4 blue beads means =1^4=1 3 green beads means=2^3=8 2 yellow beads means=3^2=9 1 orange beads means=4^1=4 product = 1x8x9x4=288 /5 yields r= 3 therefore yellow ..(D)



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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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23 Mar 2016, 22:36
kzivrev wrote: I think this question should be more precise and mention that the numbers that are marked are actualy the reminders. To me was not clear what the author wants to say. does anyone else had this issue? Hi, The Q says two things 1) The number marked on each bead is a distinct positive number and it is not the remainder.. 2) the colour on the bead is dependent on the remainder as given in the Q..Hope it clears the query
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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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02 Apr 2016, 07:12
anudeep133 wrote: Bunuel wrote: Each bead in an urn is marked with a distinct positive integer and colored according to that integer’s remainder after division by 5, as shown in the following table:
Remainder Color 0  Red 1  Blue 2  Green 3  Yellow 4  Orange
Four blue beads, three green beads, two yellow beads, and one orange bead are withdrawn. If the product of the numbers on these beads is displayed on another bead, according to the rules above, then the color of that bead is…
(A) red (B) blue (C) green (D) yellow (E) orange
Kudos for a correct solution. Solution : Red beads will be of the form 5k, blue  5k+1 , green  5k+2 , yellow  5k+3, orange  5k+4 . So, product will be (5k1+1)(5k2+1)(5k3+1)(5k4+1)(5k5+2)(5k6+2)(5k7+2)(5k8+3)(5k9+3)(5k10+4) Which will effectively be 5(p) + (1)(1)(1)(1)(2)(2)(2)(3)(3)(4) = 5(p) + (1)(8)(9)(4) = 5(p) + 8 = 5(p) + 3. Therefore, Yellow.
Otherwise, just put the values 5 for red, blue  1 , green  2 , yellow  3, orange  4. Product = (1)(1)(1)(1)(2)(2)(2)(3)(3)(4) = 288 = remainder 3. Yellow. Option D. Can you please explain how you got the product > 5(p) + (1)(1)(1)(1)(2)(2)(2)(3)(3)(4) = 5(p) + (1)(8)(9)(4) = 5(p) + 8 = 5(p) + 3 from (5k1+1)(5k2+1)(5k3+1)(5k4+1)(5k5+2)(5k6+2)(5k7+2)(5k8+3)(5k9+3)(5k10+4) Thank you for your help



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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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02 Apr 2016, 07:38
MeghaP wrote: anudeep133 wrote: Bunuel wrote: Each bead in an urn is marked with a distinct positive integer and colored according to that integer’s remainder after division by 5, as shown in the following table:
Remainder Color 0  Red 1  Blue 2  Green 3  Yellow 4  Orange
Four blue beads, three green beads, two yellow beads, and one orange bead are withdrawn. If the product of the numbers on these beads is displayed on another bead, according to the rules above, then the color of that bead is…
(A) red (B) blue (C) green (D) yellow (E) orange
Kudos for a correct solution. Solution : Red beads will be of the form 5k, blue  5k+1 , green  5k+2 , yellow  5k+3, orange  5k+4 . So, product will be (5k1+1)(5k2+1)(5k3+1)(5k4+1)(5k5+2)(5k6+2)(5k7+2)(5k8+3)(5k9+3)(5k10+4) Which will effectively be 5(p) + (1)(1)(1)(1)(2)(2)(2)(3)(3)(4) = 5(p) + (1)(8)(9)(4) = 5(p) + 8 = 5(p) + 3. Therefore, Yellow.
Otherwise, just put the values 5 for red, blue  1 , green  2 , yellow  3, orange  4. Product = (1)(1)(1)(1)(2)(2)(2)(3)(3)(4) = 288 = remainder 3. Yellow. Option D. Can you please explain how you got the product > 5(p) + (1)(1)(1)(1)(2)(2)(2)(3)(3)(4) = 5(p) + (1)(8)(9)(4) = 5(p) + 8 = 5(p) + 3 from (5k1+1)(5k2+1)(5k3+1)(5k4+1)(5k5+2)(5k6+2)(5k7+2)(5k8+3)(5k9+3)(5k10+4) Thank you for your help Hi, (5k1+1)(5k2+1)(5k3+1)(5k4+1)(5k5+2)(5k6+2)(5k7+2)(5k8+3)(5k9+3)(5k10+4).. when you will open the brackets, all terms will be multiple of 5 apart from the PRODUCT of the terms 1,1,1,1,2,...,4example(5k3+1)(5k4+1)(5k5+2)= 5k3(5k4+1)(5k5+2)+ 1(5k4+1)(5k5+2)= 5k3(5k4+1)(5k5+2)+ 1{5k4(5k5+2)+1(5k5+2)} => 5k3(5k4+1)(5k5+2)+ 1*5k4(5k5+2)+1*1(5k5+2) = 5k3(5k4+1)(5k5+2)+ 1*5k4(5k5+2)+1*1*5k5+1*1*2 So all terms shown in highlighted portion are multiple of 5.. so 5 is taken out as common and rest are given a variable p..so 5p + 1*1*2.. since we are concerned only about remainders after div by 5 Hope it cleared your doubt
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Re: Each bead in an urn is marked with a distinct positive integer and col [#permalink]
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27 Nov 2016, 23:02
Bunuel wrote: Each bead in an urn is marked with a distinct positive integer and colored according to that integer’s remainder after division by 5, as shown in the following table:
Remainder Color 0  Red 1  Blue 2  Green 3  Yellow 4  Orange
Four blue beads, three green beads, two yellow beads, and one orange bead are withdrawn. If the product of the numbers on these beads is displayed on another bead, according to the rules above, then the color of that bead is…
(A) red (B) blue (C) green (D) yellow (E) orange
Kudos for a correct solution. \(\frac{1^4*2^3*3^2*4}{5}\) \(2^2 = 4 = 1(mod 5\)), \(3^2 = 9 = 1 (mod 5)\) \(\frac{1*(1)*2*(1)*(1)}{5} = \frac{2}{5}\)\(= 2 (mod 5)\) \(= 3 (mod 5)\) Yellow D.



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