ISI2025-MCS-PCB (Non-CS) | Question: 2

Question

1. Let $b_{n} b_{n-1} \cdots b_{1}$ be the decimal representation of an $n$ digit number $m$. Let $b_{n} b_{n-1} \cdots b_{2}$ be the integer $a$ obtained from $m$ by stripping off the unit's digit $b_{1}$. Then, show that $m$ is divisible by $7$ if and only if $a-2 b_{1}$ is divisible by $7$.
2. Prove that:

\[
1+\binom{1001}{1}+\binom{1002}{2}+\cdots+\binom{2023}{1023}+\binom{2024}{1024}=\binom{2025}{1024}
\]

Answer 1

a)

$b_nb_{n-1}....b_1 = 10^{n-1}b_n + ...+ b_1 = m$

$b_nb_{n-1}...b_2 = 10^{n-2}b_n +...+ b_2 = a$

$\therefore 10a + b_1 = m$

$If \ a -2b_1 \equiv 0\ (mod \ 7) \ then  a = 2b_1 + 7k $

$=> 10a = 20b_1 + 70k $

$=> 10a + b_1 = 21b_1 + 70k = 7(3b_1 + 10k) $ which is divisible by 7 (Forward direction)

 

$ If \ 10a + b_1 = 7l $

$=> 3(a-2b_1) + 7a +7b_1 = 7l $

$=> 3(a-2b_1) = 7(l-(a+b_1)) $

$=> 3(a-2b_1)$ is divisible by 7, ie $a-2b_1$. is divisible by 7 (Backward direction)

Hence Proved