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Number Theory Problem

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by raunakrajan » Tue Jul 13, 2010 7:21 am
if (x # y) represents the remainder that results when the positive integer x is divided by the positive integer y, what is the sum of all the possible values of y such that (16 # y) = 1?

(A)8

(B)9

(C)16

(D)23

(E)24
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by amising6 » Tue Jul 13, 2010 7:27 am
raunakrajan wrote:if (x # y) represents the remainder that results when the positive integer x is divided by the positive integer y, what is the sum of all the possible values of y such that (16 # y) = 1?

(A)8

(B)9

(C)16

(D)23

(E)24
16#y=1
when 16/y you get remainder 1
we can assume 16=n*y+1
15=n*y
so find all the factor of 15
1*15
3*5
so if you divide 16 by 3 ,5 and 15 you will get remainder 1
so sum=3+5+15=23
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by kvcpk » Tue Jul 13, 2010 7:28 am
raunakrajan wrote:if (x # y) represents the remainder that results when the positive integer x is divided by the positive integer y, what is the sum of all the possible values of y such that (16 # y) = 1?

(A)8

(B)9

(C)16

(D)23

(E)24
(16 #15) = 1
(16 # 5) =1
(16 # 3) =1
15+5+3 = 23.

Bottom line is 16 will give reminder 1 when divided by any factor of 15. [except 1]

Hope this helps!!
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