How many 6 digit numbers?
Number Theory
Level
2
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.
PRE- INDIAN REGIONAL MATHEMATICAL OLYMPIAD – 2018
The answer is 70.
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1 solution
abccba can be written as
= a(10^5 + 1) + b(10^4 + 10) + c(10^3 + 10^2)
This can be written as a multiple of 7 + reminder . For the expression to be divisible by 7 the reminder should also be divisible by 7.
= (14286 * 7* a) -a + (1430 * 7 * b) + 157 7 c + c
= 7k + (c-a)
So for a(10^5 + 1) + b(10^4 + 10) + c(10^3 + 10^2) to be divisible by 7 it is sufficient if c-a is divisible by 7.
As c and a lie between 0 and 9, c-a is divisible by 7 if c-a = 7, 0 or -7
This can hence happen when ( c = 9 , a= 2) , (c=8, a=1) or (c=2, a=9) or (c=9 or a = 2) or when (a= 7 , c = 0) , the case when a = 0 and c = 7 cannot be considered as a is the starting digit of the 6 digit number and so cannot be 0. The remaining possibilities are when c = a and a not equal to 0 which leaves us with 9 cases.
So there are a total of 9 +5 = 14 possibilities can occur , in addition as b can be only odd leaves us with 14* 5 = 70 numbers which would be divisible by 7 and which would meet the required conditions stated in the problem