Math Competition Problems - Problem of the week - 4
Updated: Oct 18, 2020
"Janine's number has 3 digits. One digit is a prime number; another is a square number; the third is neither prime nor square. The number is NOT divisible by 3. What is the greatest possible value of Janine's number?"
Source - MOEMS 2008 (Grades 4-6)
Let's first list all the prime numbers, square numbers and the other numbers (neither prime nor squares)
Prime numbers: 2, 3, 5, 7
Square numbers: 1, 4, 9
Neither prime nor squares: 6, 8
To get the greatest 3 digit number the hundreds place has to be the highest.
So 9, the square number takes the hundred's place.
The next highest number is 8 (neither prime nor square) which takes the tens place.
And we can have 7 (prime number) in the ones place.
So the number can be 987. But the number should NOT be divisible by 3. And 987 is divisible by 3.
So we take the next highest number by changing the ones digit. We go for 985 (the next highest in the prime numbers). 985 is not divisible by 3. Hence the greatest possible value of Janine's number is 985.
These kind of questions become so much easier if you make an ordered list. List all your options and then pick and chose. Now wasn't that easy!
Did you enjoy the problem? Would you like to get these problems along with solutions delivered to your inbox every week? Click here to subscribe
#Competition, #Math, #Problems #MathCompetitionProblems, #ProblemOfTheWeek, #OnlineMathTutor, #OnlineMathTutoring, #LogicalReasoning, #DivisibilityTests, #NumberTheory, #OrderedList
Previous week's problem Next week's problem