Question

ANYONE can answer this??????????????????


Let n be a 5 digit number. When n is divided by 100 , let q be the quotient and r be the remainder . For how many values of n does : 11 | q + r

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Answer 1

That means we have,

N=100q+r

Substract 99q from both sides to obtain,

(N-99 q) = q+r

Now, we need that q+r to be divisble by 11.
Because of equality it is equivalent to saying,
That we need (N- 99q) to be divisible by 11.
But 99q is already divisible by 11.
Thus, we require that N be divisible by 11.

Thus, what the question is really asking is how many 5 digits numbers are divisible by 11.

Let us list all 5 digit numbers in increasing order.
10000
10001
10002
10003
.....
99999

The fist number on this list divisible by 11 is:
10010
And after that each 11-th number is divisble of 11.
Thus,
10010
10021
10032
....
Up to 99999

Now, I leave if to you to figure out how many numbers are in that list.


First find the last number in the range 10000-99999
Divisible by 11.
Start working backwards from 99999.
We soon find that 99990 is the last such number. 

Thus, how many number are in here,
10010
10021
10032
10043
....
99990

This is an arithmetic series, who constant difference is 11.
Thus, the pattern has the form,
(n-th term) = 11*n+K
Where K is the konstant to be determined.

The first term is 10010, thus,
(1st term) = 11(1)+K
10010=11+K
K=9999

Thus,
(n-th term) = 11*n+9999

For example,
(4-th term) = 11(4)+9999 = 10043
Which matches with the number on the last.

Thus, to determine which number in the sequence, 99990 is we need to solve,
99990=11*n+9999
Thus, n=8181

Thus, 99990 is the 8181-st term in the sequence.
Thus, there are 8181 terms.

Answer 2

if q is quotient and r is remainder  

then n = 100q + r = 99q + (q+r)  

if q+r is divisible by 11 then n is divisible by 11  

so our question boils down to how many 5 digit numbers are divisible by 11  

lowest 5 digit number = 10000 = 11 * 909 + 1  

largest 5 figit number = 99999 = 9090*11 +9  

number of numbers divisible by 11 = 9090 - 909 = 8181