Question

consider all 6 digit number of the form abccba, where b is odd. determine the number of all such 6 digit number that are divisible by 7

Answer 1

Yesterday I have Also Given PRMO
but I got 80 but In answer key 70 numbers were given...I don't know whether PRMO Key is correct or not

Using divisibility rule of 7
3a+2b-c-3c-2b+a
4(a-c)

for a 6digit number to be divisible by 7
4(a-c) should be divisible by 7

for this to happen (a-c) should be 7 , -7 , 0

therefore number of pair we get satisfying this realtion would be...(9-2) , (8-1) , (7-0) , (2-9) , (1-8) , (0-7) , (0-0) , (1-1) , (2-2) , (3-3) , (4-4) , (5-5) ,(6-6) , (7-7) , (8-8) , (9-9)
Total pair = 16

but B = 1 , 3 , 5 , 7 ,9

Therefore Number of possible 6 digit No. can be made with 16 pair and 5 numbers
is 16 × 5 = 80

Anyone can correct me if you find something wrong in it...

I Know PRMO Key mention it 70 as answer

Answer 2

It is a case of generating 6 digit palindrome numbers abccba under two constraints,

- b is 1, 3, 5, 7 or 9

- the number is divisible by 7

If (a - c) = 0 or 7, that number is divisible by 7.

SEE the attachments for the detailed explanation and the list of 70 palindromes divisible by 7.