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Can anyone teach me how to do this question? Thanks.
Date Posted: 5 years ago
Views: 429
17/10
= 1 + 7/10
= 1 + 1/(10/7)
= 1 + 1/(1 + 3/7)
= 1 + 1/[1 + 1/(7/3)]
= 1 + 1/[1 + 1/(2 + 1/3)]
So a is 1, b is 2, c is 3
Their sum is 6.
= 1 + 7/10
= 1 + 1/(10/7)
= 1 + 1/(1 + 3/7)
= 1 + 1/[1 + 1/(7/3)]
= 1 + 1/[1 + 1/(2 + 1/3)]
So a is 1, b is 2, c is 3
Their sum is 6.
Thank you very much, Eric!
Alternative method :
1 + 1/(a + 1/(b + 1/c) )
= 1 + 1/[ a + 1/((bc+1)/c) ]
= 1 + 1/[a + c/(bc+1)]
= 1 + 1/[(abc + a + c)/(bc+1)]
= 1 + (bc+1)/(abc + a + c)
17/10 = 1 + 7/10
Comparing the terms,
bc + 1 = 7
bc = 6
abc + a + c = 10
Sub bc = 1, 6a + a + c = 10
c = 10 - 7a
Since a and c are positive integers, only possible value of a is 1.
c = 10 - 7(1) = 3
Since bc = 6 , b = 6/3 = 2
a + b + c = 1 + 2 + 3 = 6
1 + 1/(a + 1/(b + 1/c) )
= 1 + 1/[ a + 1/((bc+1)/c) ]
= 1 + 1/[a + c/(bc+1)]
= 1 + 1/[(abc + a + c)/(bc+1)]
= 1 + (bc+1)/(abc + a + c)
17/10 = 1 + 7/10
Comparing the terms,
bc + 1 = 7
bc = 6
abc + a + c = 10
Sub bc = 1, 6a + a + c = 10
c = 10 - 7a
Since a and c are positive integers, only possible value of a is 1.
c = 10 - 7(1) = 3
Since bc = 6 , b = 6/3 = 2
a + b + c = 1 + 2 + 3 = 6
I had attempted this question with your method at first but didn't know how to solve the equation. Thank you very much J for your kind input!
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Answer as shown in pic
Date Posted:
5 years ago
Thank you very much Dawn! Really appreciate your help!
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