Question

Let n be the no. Of numbers divisible by 8 out of all the 4-digit numbers that can be formed with digits 4,5,6,7,8 no digits being repeated. What is the value of n ?

Answer 1

By writing the digits as \(abcd := a*10^3+b*10^2+c*10+d\) we can look at what it will be in mod 8, since 0 in mod 8 means divisible by 8. In doing so we can replace the powers of 10 with powers of 2:

\[a*10^3+b*10^2+c*10+d \mod 8\]\[a*2^3+b*2^2+c*2+d \mod 8\]\[b*2^2+c*2+d \mod 8\]

Interestingly, the leading digit can be any number and it will always be divisible by 8 as long as the next 3 digits are congruent to 0 mod 8.

Answer 2

Why does 0 in mod 8 mean divisible by 8 also could you explain again why you replaced the powers of 10 with 2 ?

Answer 3

Ah, well mod 8 means you take the remainder when you divide something by 8. So if you divide something by 8 and you get 0, that means there's no remainder. In other words, it's divisible by 8!

Similarly we can replace 10 with 2 because 2 is the remainder of 10 when you divide it by 8.

Answer 4

To add a little something to the last sentence of @Kainui example: \(a \times 10^3\) is replaced with \(a \times 2^3\) because, after writing \(10=2+8\), you obtain
\[ (2+8)^3 = 2^3 + 3\dots 8 + 3\dots 8^2+ 8^3.\] All terms are multiples of 8 except \(2^3\).

Answer 5

Thank you so much guys ! Got it :)