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Re M2627
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16 Sep 2014, 01:26
Official Solution:\(x\), \(y\) and \(z\) are positive integers such that when \(x\) is divided by \(y\) the remainder is 3, and when \(y\) is divided by \(z\), the remainder is 8. What is the smallest possible value of \(x+y+z\)? A. 12 B. 20 C. 24 D. 29 E. 33 Given \(x=qy+3\), where \(q\) is a quotient, an integer \(\ge 0\). Which means that the least value of \(x\) is when \(q=0\), in that case \(x=3\). This basically means that \(x\) is less than \(y\). For example 3 divided by 4 yields the remainder of 3. According to the above: \(x\) is divided by \(y\) the remainder is 3 means that the minimum value of \(x\) is 3; \(y\) is divided by \(z\) the remainder is 8 means that the minimum value of \(y\) is 8 and the minimum value of \(z\) is one more than 8, so 9 (8 divided by 9 yields the remainder of 8); So, the smallest possible value of \(x+y+z\) is \(3+8+9=20\). Answer: B
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Re: M2627
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08 Jun 2015, 21:58
wow I guessed this one but its a brilliant problem! Beautiful!



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05 Dec 2015, 02:18
if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4



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Re: M2627
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16 Sep 2016, 01:29
please explain.. if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4 please can anyone explain me this problem in easy way



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16 Sep 2016, 02:04



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Karanagrawal wrote: please explain.. if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4 please can anyone explain me this problem in easy way See x has to be at least 3 for a remainder of 3 when x is divided by y. then x = 3. y has to be greater than 3 for x/y to get a remainder of 3, and y has to be at least 8 for y/z to get a remainder of 8. then y = 8. z has to be greater than 8 for y/z to have a remainder of 8. Lowest possible integer greater than 8 is 9. So going with z = 9. x + y + z = 3 + 8 + 9 = 20
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Re: M2627
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16 Sep 2016, 02:43
Karanagrawal wrote: please explain.. if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4 please can anyone explain me this problem in easy way Remember this rule: If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
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Karanagrawal wrote: please explain.. if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4 please can anyone explain me this problem in easy way Hi, X as 2 and y as 4 is clearly incorrect... Now for this Q, my advice would be talk of larger number first... 1) If y is divided by z, remainder is 8.. This tells us that least value of y is 8 and the remainder will be 8, ONLY if z is greater than 8, so take the next least possible value.. And it is 8+1=9... 2) now take the second part.. If x is divided by y, remainder is 3.. Since least value of y is 8, x will be 3 or 11 or 19 and so on to leave remainder of 3.. But we have to take least value and it is 3.. So x can be 3, y can be 8 and z can be 9.. Ans x+y+z=3+8+9=20
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Re: M2627
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23 Mar 2017, 11:18
deospaima wrote: if x divided by y the reminder is 3 then y > 3 , not x . example : x=2 ; y=4 as stated minimum value of x=3 and y>3 it means for minimum value of X y>x see eg: x=3 and y=4 this means when x is divided by y remainder is 3 and y>remainder i.e. 3 but in this case y if >3 will also be greater than x ultimately.



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12 Aug 2017, 08:12
I think this is a highquality question and I agree with explanation.



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Re: M2627
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18 Oct 2017, 16:31
Would this method be wrong?
x/y > remainder = 3 therefore y = x+3
y/z > remainder of 8 therefore z = x+ 8
x+ 8 + x + x+ 3 = 11 + 3x  the least multiple of 3 that results in any answer is 3
therefore 3*3 + 11 = 20



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18 Oct 2017, 21:13



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Re: M2627
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24 Dec 2017, 04:48
Given: x=y(integer) + 3
y=z(integer) + 8
Therefore, x=z(integer) +8 + 3
x + y + z = [z(integer) +8 + 3] + [z(integer) + 8] + z = z(0) + 8 + 3+ z(0) + 8 + z = 19 + z
Minimum possible value of z = 1
thus, x + y + z = 19 + 1 = 20



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Hey BunuelIf x/y leaves a remainder of 3 and y/z leaves a remainder of 8, how can I say that y is 4? I get that 3/4 leaves a remainder of 3 which complies with the first part of the question, but (y=4)/z can not leave a remainder of 8. So, if y/z leaves a remainder of 8, then y=8 and z=9 and with x/y leaving a remainder of 3, shouldn't x=8+3=11? But the biggest problem is, the options do not have 8+9+11=28 as an option.
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28 Aug 2018, 10:00
I kept wondering about this question and I figured it out finally. Even with the denominator of 8, 3 could be the remainder if the numerator is 3. Thus x + y +z = 3 + 8 + 9 = 20 Therefore B: 20
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