 |
|
|
|
|
SCHOOL SUBJECTS
-
-
-
-
-
-
-
-
-
-
-
-
|
|
|
|
GRADE LEVELS
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
|
|
|
|
|
Post a New Question | Current Questions | Chat With Live Tutors
Homework Help Forum: math
Posted by lony on Monday, August 31, 2009 at 4:38am.
|
the digits of a certain number of 3 arithmetical progression.Their sum is 15 and the number is 129 times its first digit.What is the number?
|
- math - Reiny, Monday, August 31, 2009 at 8:21am
Something is not right about your question.
First of all the first sentence is not complete.
Did you mean "The digits of a certain number of 3 digits form an arithmetical progression." ?
When you say, "the number is 129 times the first digit", do you mean the unit digit or the hundredth digit.
either way, 129 = 3*43, and since 43 is prime, either the unit digit is 3 or the hundredth digit is 3.
now the sum of the digits has to be 15, and to have the numbers in an arithmetic sequence, the only possible numbers to satisfy both conditions are :
357 or 753
But neither of these two satisfies the condition of being 129 times the first digit.
BTW, I also tried this:
let the first digit be a
the second digit be a+d
the third digit be a+2d
then forming two equations in 2 unknowns using both interpretations.
In both cases, the values of a did not come up as a positive integer, as it should have.
|
Answer this Question
|
|
|
|
