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
Let for all n , natural numbers f(n) = product of non zero digits of n.find the largest prime divisor of
f(1) + f(2) + ...... + f(999). (the answer is 103)
The question is the largest prime no that divides f(n)
Lets simplify f(1)+f(2)+f(3)+f(4)+...+f(999)
= (1+2+3+4...+9)+1+(1+2+3+4+...+9)+2+2(1+2+3+4+...+9) +3+3(1+2+3+4+...+9) +...
On simpifying and using formula for summation of n natural no n(n+1)/2we get that the summation is
Full solution is not given as u needed hints fr the question
((9*10)/2+1)((9*10)/2+1)((9*10)/2+1) -1 = (46*46*46)-1 97335
On prime factorization os 97335 we getthe last prime as 103 and hence we said tht the largest prime tht devides
Thnks for ur reply
But,what does product of non zero digits of n
product of non zero digits of n
mean?
The product of non zero digits of n is like f(999) = 9*9*9 , f(998)= 9*9*8, f(101)= 1*1 (0 not included)
How did u do this?
f(1)+f(2)+f(3)+f(4)+...+f(999)= (1+2+3+4...+9)+1+(1+2+3+4+...+9)+2+2(1+2+3+4+...+9) +3+3(1+2+3+4+...+9) +...
This can be done in following way
This can be done in f(1)= 1, f(2)=2, ... ,f(9)=9 , f(10) = 1 , f(11) =(1*1), f(12)= (1*2), f(13)=(1*3),...f(19),f(20)= 2 , f(21)= (2*1) and so on
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 !