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by sanju09 » Mon Mar 08, 2010 5:26 am
What is the greatest number, which will divide 215, 167, and 135 so as to leave the same remainder in each case?
(A) 16
(B) 18
(C) 24
(D) 32
(E) 64
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by ajith » Mon Mar 08, 2010 5:32 am
sanju09 wrote:What is the greatest number, which will divide 215, 167, and 135 so as to leave the same remainder in each case?
(A) 16
(B) 18
(C) 24
(D) 32
(E) 64
say r is the remainder and k is the number

215 = a*k +r
167 = b*k+ r
135 = c*k +r

(167-135) = (b-c)*k
32 = (b-c)*k
48 = (a-b)*k
80 = (a-c)*k

since k has to be the maximum k will be the LCM of 32,48 and 80 which is 16

A
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by sanju09 » Mon Mar 08, 2010 5:36 am
ajith wrote:
sanju09 wrote:What is the greatest number, which will divide 215, 167, and 135 so as to leave the same remainder in each case?
(A) 16
(B) 18
(C) 24
(D) 32
(E) 64
say r is the remainder and k is the number

215 = a*k +r
167 = b*k+ r
135 = c*k +r

(167-135) = (b-c)*k
32 = (b-c)*k
48 = (a-b)*k
80 = (a-c)*k

since k has to be the maximum k will be the LCM of 32,48 and 80 which is 16

A
sure! :?:
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by rohan_vus » Mon Mar 08, 2010 5:42 am
ajith wrote:
sanju09 wrote:What is the greatest number, which will divide 215, 167, and 135 so as to leave the same remainder in each case?
(A) 16
(B) 18
(C) 24
(D) 32
(E) 64
say r is the remainder and k is the number

215 = a*k +r
167 = b*k+ r
135 = c*k +r

(167-135) = (b-c)*k
32 = (b-c)*k
48 = (a-b)*k
80 = (a-c)*k

since k has to be the maximum k will be the LCM of 32,48 and 80 which is 16

A
I guess you meant you say GCF and not LCM
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by ajith » Mon Mar 08, 2010 5:50 am
rohan_vus wrote:
I guess you meant you say GCF and not LCM
Thanks Rohan and Sanju, Yes Indeed! Thankfully it is just a name and I do not need to alter the calculations
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by kstv » Tue Mar 09, 2010 10:41 am
What is the greatest number, which will divide 215, 167, and 135 so as to leave the same remainder in each case?

215 - 167 = 48
167- 135 = 32

G C F of 48 and 32 is 16

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