Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
If a b c are three positive intejers{I}.Such that ab+bc+ac+a+b+c=989
then find a+b+c=?
Dear student,
abc+ab+ac+bc+a+b+c=989, abc+ab+ac+bc+a+b+c+1=990 (a+1)(b+1)(c+1)=990 Since a,b,c >0, then (a+b+c+3)/3= ((a+1)+(b+1)+(c+1))/3>= ((a+1)(b+1)(c+1))^(1/3)=(990)^(1/3) >(729)^(1/3)=9 So a+b+c+3>27, a+b+c>24. 990=2*(3^2)*5*11 Since a,b,c are all positive integers, hence >0, then (a+1), (b+1), (c+1) are all >=2. We are not given that the sought numbers are consecutive, this makes the problem a little lengthy. Then to mention a few, the following options are possible: a+1=2, b+1=9, c+1=55 => a=1, b=8, c=54. Then a+b+c=63 a+1=2, b+1=3, c+1=165 => a=1, b=2, c=164. Then a+b+c=167, a+1=9, b+1=10, c+1=11 => a=8, b=9, c=10, Then a+b+c=27 a+1=2, b+1=5, c+1=99 => a=1, b=4, c=98 =>a+b+c=103 and so forth in a finite number of steps all relevant a,b,c with a+b+c>24 can be listed
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !