ISI2025-MCS-PCB (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 $b_{n} b_{n-1} \ldots b_{2}$ 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)  $m = b_{n}b_{n-1} \cdots b_1$

      $a = b_{n}b_{n-1} \cdots b_2 $

we can write $m$ in terms of $a$ as follows:

$m = 10\times a  + b_1$

let's manipulate it a little bit,

$ m = 10a + b_1 = 10a -20b_1 + 21b_1 = 10(a - 2b_1) + 21b_1$

$\implies  m = 10(a - 2b_1) + 21b_1$, 

hence, if $m$ is divisible by $7$, then  $(a - 2b_1)$ must be divisible by $7$.

 

Now let's prove it around the way, 

$m = 10(a - 2b_1) + 21b_1 \implies m - 21b_1 = 10(a - 2b_1)$

$\implies a - 2b_1 = \frac{m - 21b_1}{10}$,

So, if $a-2b_1$ is divisible by $7$, then $m$ must be divisible by $7$. 

Hence, we conclude $m$ is divisible by $7$ if and only if $a-2b_1$ is divisible by $7$.

 

---

b)

we know that $\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$

$\implies \binom{n}{k} = \binom{n+}{k}  \binom{n}{k-1}$

Apply the above to each term of the LHS, then we get

LHS =

$1+\binom{1001}{1}+\binom{1002}{2}+\cdots+\binom{2023}{1023}+\binom{2024}{1024}$

= $1+\left [ \binom{1002}{1} - \binom{1001}{0} \right ] + \left [ \binom{1003}{2} - \binom{1002}{1} \right ]+\cdots+ \left [ \binom{2024}{1023} - \binom{2023}{1022} \right ]+ \left [ \binom{2025}{1024} - \binom{2024}{1023} \right ]$

Each negative term cancels the positive term of the previous [ ], we are left with 

$= 1 - \binom{1001}{0} + \binom{2025}{1024} $

because $\binom{n}{0} = 0$, we have,

$= 1 - 1 + \binom{2025}{1024} = \binom{2025}{1024}$

= RHS.