Try this beautiful problem from the PRMO, 2018 based on Digits of number.
Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7.
Algebra
Numbers
Multiples
But try the problem first...
Answer: is 70.
PRMO, 2018, Question 3
Higher Algebra by Hall and Knight
First hint
abccba (b is odd)
=a(\(10^5\)+1)+b(\(10^4\)+10)+c(\(10^3\)+\(10^2\))
=a(1001-1)100+a+10b(1001)+(100)(11)c
=(7.11.13.100)a-99a+10b(7.11.13)+(98+2)(11)c
=7p+(c-a) where p is an integer
Second Hint
Now if c-a is a multiple of 7
c-a=7,0,-7
hence number of ordered pairs of (a,c) is 14
Final Step
since b is odd
number of such number=\(14 \times 5\)=70.
Try this beautiful problem from the PRMO, 2018 based on Digits of number.
Consider all 6-digit numbers of the form abccba where b is odd. Determine the number of all such 6-digit numbers that are divisible by 7.
Algebra
Numbers
Multiples
But try the problem first...
Answer: is 70.
PRMO, 2018, Question 3
Higher Algebra by Hall and Knight
First hint
abccba (b is odd)
=a(\(10^5\)+1)+b(\(10^4\)+10)+c(\(10^3\)+\(10^2\))
=a(1001-1)100+a+10b(1001)+(100)(11)c
=(7.11.13.100)a-99a+10b(7.11.13)+(98+2)(11)c
=7p+(c-a) where p is an integer
Second Hint
Now if c-a is a multiple of 7
c-a=7,0,-7
hence number of ordered pairs of (a,c) is 14
Final Step
since b is odd
number of such number=\(14 \times 5\)=70.
