Help Solve the Maths Challenge: Positive Integer with All Digits 1

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Homework Help Overview

The problem involves finding a positive integer X, composed entirely of the digit 1, such that the expression pX^2 + qX + r results in a number where all digits are also 1. The coefficients p, q, and r are fixed integers, with p being greater than 0. The challenge is to determine a possible value for q under these conditions.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss substituting specific values for X and p to explore potential values for q. There is a suggestion to analyze the implications of X being a number with varying digits of 1.
  • Some participants question whether the conditions apply universally to all permutations of X, considering the fixed nature of p, q, and r.
  • One participant muses about the structure of X^2 and its implications for the digits of the resulting expression, noting the potential complexity of the digits involved.
  • Another participant raises the idea of specific values for q and r, questioning their impact on the overall expression.

Discussion Status

The discussion is ongoing, with various interpretations and approaches being explored. Some participants are attempting to clarify the mathematical relationships involved, while others are questioning the assumptions made about the structure of X and its implications for the coefficients. There is no explicit consensus yet, but several lines of reasoning are being examined.

Contextual Notes

Participants note the constraints of the problem, including the requirement that the expression must hold true for any number of digits in X. There is also a mention of the lack of familiarity with number theory among some participants, which may affect their contributions to the discussion.

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one of the "maths challenge" questions. one of the only ones i couldn't do :S I've tried for a good number of hours, but havnt found a way to do it. I feel any more time would just be brute forcing it which i wouldn't have time for in the exam anyways.

hope you can help

Homework Statement



X is a positive integer in which each digit is 1; that is, X is of the form 111111...
Given that every digit of the integer pX^2 + qX + r (where p, q, and r are fixed integer coefficients and p > 0) is also 1. Irrespective of the number of digits X. Which of the following is a possible value of q?


Homework Equations





The Attempt at a Solution



:S


thanks
 
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Well if I understood correctly. Then can't you substitute x=1 and p=1 and then see what numbers for q would give you an integer with each digit being 1. Understand?
 
it has to work for every possibly permutation of X I think.

hence "Irrespective of the number of digits X" and "where p, q, and r are fixed".

is that right?
 
I know absolutely zero number theory, but here are my musings on this:

If X=11111... with n 1's, then X^2 contains all the digits from (and contained in) 1 to n, no more and no less.So
X^2 = 123...n...321 where n is the no. of 1's in X for n<11 (10=>0). n>10, X^2 contains all numbers 0, 1, ...9
pX^2 = p(123...n...321)
qX = qqqqq... n q's in total

So q=0 if n=1 and r= 0. For higher n the digits of X^2 are not all 1's.

p(123...n...321) + qqqq...+r = 111...
Is there a digit q that satisfies:
p(123...n...321) + qqqq... +r = 1111...

Apparently not: p(123...n...321) is of length 2n-1 (number of digits it contains), and qqqq... is of length n.
Im obviously not sure of all this, but it seems like this may have something to do with it. If it does, you can work it out and formalise it, if not, you can ignore me :D
 
Last edited:
well if q is 10 and r is 0 won't the integer be 11?

Edit: qspeechc might be right in what he's saying but I can't understand it at the moment but I hope you did.
 
Last edited:
To repeat myself; I don't know any number theory.

If q is 10^n, then all it does is add n zeroes on the end of the 1's, and you can see then that it doesn't change anything.
 
I don't think he's checked this.
 

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