Solve: Consider $f (x) = 90x^2 + 20x + 1$ then sum of digits of $f (111111)$ is...?

Question

This is the question asked in my maths paper of quadratic equations but I am unable to understand which concept will be used here . Please help me in this.

Just to note: $111111^2 =12345654321$ – Berci Sep 21 '18 at 07:42 — Berci, Sep 21 '18 at 07:42

score 2 · Accepted Answer · edited Sep 21 '18 at 08:36

2

I have calculated the answer. It is 13. $f(1)=111$, sum of digits$=3$.

$f(11)=121*90+20*11+1=11111$, sum of digits$=5$.

In same way when you calculate this sum of digits of

$f(111)=7$

$f(1111)=9$

$f(11111)=11$

$f(111111)=13$

Your answer. The form written is the trick. ($90x^2+20x+1$)

edited Sep 21 '18 at 08:36

Henrik supports the community

6,831

answered Sep 21 '18 at 07:55

rohan

36

Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. – José Carlos Santos Sep 21 '18 at 08:01

farruhota · Answer 2 · 2018-09-21T08:13:08.250

1

Hint: Complete the square: $$90x^2+20x+1=90(x+\frac19)^2-\frac19=\\ =\frac{10}{9}(9x+1)^2-\frac19=\frac{10(9x+1)^2-1}{9}.$$ Answer: $13$.

edited Sep 21 '18 at 08:13

answered Sep 21 '18 at 08:01

farruhota

31,482

right, $(10^{13}-1)/9$ gives the result. – farruhota Sep 21 '18 at 08:12

Solve: Consider $f (x) = 90x^2 + 20x + 1$ then sum of digits of $f (111111)$ is...?

2 Answers2