Consider the identity $(1+x)(10+x)\left(10^{2}+x\right) \cdots\left(10^{10}+x\right)=10^{a}+10^{b} x+a_{2} x^{2}+\cdots+a_{11} x^{11}.$ Comment on $a$

Question

Consider the identity $(1+x)(10+x)\left(10^{2}+x\right) \cdots\left(10^{10}+x\right)=10^{a}+10^{b} x+a_{2} x^{2}+\cdots+a_{11} x^{11}.$ We denote the largest integer lesser than ot equal to $z$ as $[x].$ What can you say about $a,b,[a],[b].$ Specifically, comment which of the following options, are the most appropriate/correct:

A.$[a]>[b],$

B.$[a]=[b]$ if $a>b$

C.$[a]<[b],$

D.$[a]=[b]$ if $a<b$

I am much confused about this problem. I intially thought bashing it with polynomials. I thought the expression $10^{a}+10^{b} x+a_{2} x^{2}+\cdots+a_{11} x^{11}$ as a polynomial. The roots of this polynomial are therefore, $-1,-10,-100,...,10^{10}.$ Applying Vieta's seemed most immediate here, and I got, $$-10^a/a_{11}-1\times -10\times\cdots\times -10^{10}=-10^{50}\implies -10^{50}a_{11}=-10^a\implies 10^{50}a_{11}=10^a.$$ Also, I observed that $$\frac{-10^{50}}{-1}+\frac{-10^{50}}{-10}+\cdots+\frac{-10^{50}}{-10^{10}}=10^b/a_{11}\implies 10^{50}+10^{49}+...+10^{40}=\frac{10^{40}(10^{11}-1)}{9}=10^b/a_{11}\implies 10^{40}(10^{11}-1)a_{11}=9.10^b\implies \underbrace{10^{40}(1...1)}_{\text{ 11 one's}}a_{11}=10^b.$$ I am now assuming no $a_i's$ are zero. So, we have, $$\frac{10^{50}a_{11}}{\underbrace{10^{40}(1...1)}_{\text{ 11 one's}}a_{11}}=\frac{10^a}{10^b}\implies \frac{10^{10}}{\underbrace{(1...1)}_{\text{ 11 one's}}}=\frac{10^a}{10^b}.$$ This is never possible if $a,b$ are integers. So, my assjmption/consideration of thinking it as polynomial was wrong. How to figure the right option?

Answer 1

To give a correct approach, I edited the errors in the mentioned method in OP and enclose it in a "quoted space" below :

I intially thought bashing it with polynomials. I thought the expression $10^{a}+10^{b} x+a_{2} x^{2}+\cdots+a_{11} x^{11}$ as a polynomial. The roots of this polynomial are therefore, $-1,-10,-100,...,10^{10}.$ Applying Vieta's seemed most immediate here, and I got, $$-10^a/a_{11}-1\times -10\times\cdots\times -10^{10}=-10^{55}\implies -10^{55}a_{11}=-10^a\implies 10^{55}a_{11}=10^a.$$ Also, I observed that $$\frac{-10^{55}}{-1}+\frac{-10^{55}}{-10}+\cdots+\frac{-10^{55}}{-10^{10}}=10^b/a_{11}\implies 10^{55}+10^{54}+...+10^{45}=\frac{10^{45}(10^{11}-1)}{9}=10^b/a_{11}\implies 10^{45}(10^{11}-1)a_{11}=9.10^b\implies \underbrace{10^{45}(1...1)}_{\text{ 11 one's}}a_{11}=10^b.$$ I am now assuming no $a_i's$ are zero. So, we have, $$\frac{10^{55}a_{11}}{\underbrace{10^{45}(1...1)}_{\text{ 11 one's}}a_{11}}=\frac{10^a}{10^b}\implies \frac{10^{10}}{\underbrace{(1...1)}_{\text{ 11 one's}}}=\frac{10^a}{10^b}.$$

This is the corrected part of the OP method. For the remaining part, we proceed as:

From the equation $$10^{55}a_{11}=10^a,$$ we can say, that $a_{11}=10^s$ and hence, $a=55+s.$ Also, we have found $$\frac{10^{10}}{\underbrace{(1...1)}_{\text{ 11 one's}}}=\frac{10^a}{10^b}\implies \frac{10^{10}}{\underbrace{(1...1)}_{\text{ 11 one's}}}=10^{55+s-b},$$ this gives the feeling that $b$ can never be an integer. So, taking $\log_{10}$ on both sides, we have, $55+s-b=10-\log_{10}{\underbrace{(1...1)}_{\text{ 11 one's}}}\implies b=45+s+\log_{10}{\underbrace{(1...1)}_{\text{ 11 one's}}}.$ Now, we have the values $a=55+s$ and $b=45+s+\log_{10}{\underbrace{(1...1)}_{\text{ 11 one's}}}.$ We now, note that, $b\geq 45+s+\log_{10}{10^{10}}=55+s\geq a.$ Thus, we find $b\geq a\implies [b]\geq [a].$

Indeed, this implies the correctness of Option $C$ .