I don't know if my methods right:is it a)?
A lucky number is a positive integer which is 19 times the sum of its digits. How
many different lucky numbers are there?
Close, the actual answer is 11 :|I don't know if my methods right:
By using the conditions in the questions. It's easy to deduce that the number isn't a 1,2 or 4 digit number. So it's a 3 digit no.
Let the digits be as follow: 100 digits in hundreds place, 10 digits in 10s place and 10 digits in 1's place.
Alternatively, I think you can just go:
Z*19=ABC where {Z,A,B,C E Z}
So say for: a no. which has digits summing to 6 is:
6*19=114
Thus, Z is in between 5 and 16
So 10 numbers.
Yeah it is 11. lol. I counted my solutions in between 5 and 16 incorrectly.Close, the actual answer is 11 :|
Here is the solution:
As shaon showed, the lucky number must be a 3 digit number.
Suppose the number is abc.
Then , we have:
ie
Note that as b and c are the ten's digit and unit's digit respectively, then
So the maximum is and therefore the maximum a is
So we must only check for
For a = 1, we have
For a = 2, we have
For a = 3, we have , and there are no other solutions.
Hence there are exactly 11 lucky numbers, namely,
We should post up easier questions first just to get the marathon going =)
Hi Kurt,Here is 2008 q + a and 2006 answers.
Yes, the answer is (c)or is it (c)?
Cheers kurt! you're awesome. what do you usually get in the maths comp?
Here's a new question.
A sequence {u1, u2, . . . , un} of real numbers is defined by
for n≥ 3.
What is u2008?
(A) −√2
(B) 2008(√2 − 2008pi)
(C) 1003√2 − 1004pi
(D) pi
(E)√2