Question

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

Answer 1

abccba can be written as :

100000a + 10000b + 1000c + 100c + 10b + a

100001a + 10010b + 1100c

We can resolve it as below:

(99995a + 6a) + (10010b) + (1099c + c)

Apart fro, 6a and c all of them are multiple of 7.

Now, (6a + c) = 7a + (c - a) must be divided by7.

now c - a must be equal to either -7, 0 or 7

for c-a = -7; (0,7); (2,9); (1,8) i.e. 3 combinations.

for c-a = 0; (1,1)..... (9,9) i.e. 9 combinations.

for c-a = 7; (9,2); (8,1) i.e. 2 combinations.

for (3+9+2 = 14) b can be 1,3,5,7,9

So, 14 * 5 = 70 combinations are possible.

Answer 2

Answer:

Step-by-step explanation: a(105

+ 1) + b(104

+ 10) + c(103

+ 102

)

= 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

Now if c – a is a multiple of 7

c – a = 7, 0, –7

Hence number of ordered pairs of (a, c) is 14

since b is odd

Number of such number = 14 × 5 = 70