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#### Albert

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- Jan 25, 2013

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- Thread starter Albert
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- Jan 25, 2013

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- Mar 31, 2013

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k = c + d + g + h

k = f + g + i + j

add all 3 to get 3k = a+b+c+d+e+f+g+h+i + (d+g) = 65+ (d+g) as sum of numbers 65 + d +g is multiple of 3

for the lowest

we need to choose d+g lowest such that 65 + d +g is multiple of 3 and d + g is lowest. Do d+g mod 3 = 1 , it canntot be 1 or 4 as the lowest is 2+ 3 = 5 so 7

that gives d = 2 ; g =5 and lowest sum = 24 ( there is more than one solution and one is given below)

( a = 10, b= 8, d = 2, e = 4, c = 11, g= 5, h = 6, f = 9, i = 5, j = 3) satisfies it

for the highest

we need to choose d+g highest such that + d +g is multiple of 3 and d + g is highest . Do d+g mod 3 = 1 , it < 21 so it is 20

that gives d = 10 ; g =9 and lowest sum = 24

( a = 7, b= 3, d = 10, e = 8, c = 5, g= 9, h = 4, f = 11, i = 2, j = 6) satisfies it

Lowest k = 24, highest k = 28

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- Jan 25, 2013

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yes, your answer is correct

k = c + d + g + h

k = f + g + i + j

add all 3 to get 3k = a+b+c+d+e+f+g+h+i + (d+g) = 65+ (d+g) as sum of numbers 65 + d +g is multiple of 3

for the lowest

we need to choose d+g lowest such that 65 + d +g is multiple of 3 and d + g is lowest. Do d+g mod 3 = 1 , it canntot be 1 or 4 as the lowest is 2+ 3 = 5 so 7

that gives d = 2 ; g =5 and lowest sum = 24 ( there is more than one solution and one is given below)

( a = 10, b= 8, d = 2, e = 4, c = 11, g= 5, h = 6, f = 9, i = 5, j = 3) satisfies it

for the highest

we need to choose d+g highest such that + d +g is multiple of 3 and d + g is highest . Do d+g mod 3 = 1 , it < 21 so it is 20

that gives d = 10 ; g =9 and lowest sum = 24

( a = 7, b= 3, d = 10, e = 8, c = 5, g= 9, h = 4, f = 11, i = 2, j = 6) satisfies it

Lowest k = 24, highest k = 28