Finding a & b with Raised Digits

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Homework Help Overview

The problem involves two integers, a and b, where a is a two-digit number and b is a four-digit number, both with digits less than 9. The relationship between them is defined by the equation a^2 = b, and the problem states that if all digits of a and b are increased by one, the relationship still holds. Participants are exploring the implications of this condition.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest examining the relationship between the original numbers and their incremented forms, while others propose using specific examples to clarify the problem. There are discussions about the complexity of representing a and b in terms of their digits and the implications of the problem's wording regarding validity after digit elevation.

Discussion Status

Participants have provided various insights and approaches, including narrowing down possible values for a based on given constraints. Some have attempted algebraic manipulations to derive relationships between a and b, while others have expressed skepticism about the clarity of the problem statement. There is no explicit consensus on a single approach, but several productive lines of reasoning have emerged.

Contextual Notes

Participants note that all digits being less than 9 is a critical constraint, and there is some ambiguity regarding the term 'valid' as used in the problem statement. The discussion reflects a range of interpretations and methods being considered to tackle the problem.

dirk_mec1
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Homework Statement


Given are two integers a and b: a has 2 digits and b has 4 digits. All digits are smaller than 9. a and b are such that a^2 = b.

If all digits are raised by one this is still valid. What are a and b?

Homework Equations


N/a

The Attempt at a Solution


Suppose a = a_0 \cdot 10^0 + a_1 \cdot 10^1

and b = b_0 \cdot 10^0 + b_1 \cdot 10^1 + b_2 \cdot 10^2 + b_3 \cdot 10^3

then it gets messy if I use a^2 = b and if I increase all digits by one.

I think should get two equations with 2 unknowns.
 
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Why don't you try to find the relationship between the numbers whose digits have been increased by one and the originals ( for both 2 and 4 digits?)
Take specific examples to help if you want to.
 
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dirk_mec1 said:

Homework Statement


Given are two integers a and b: a has 2 digits and b has 4 digits. All digits are smaller than 9. a and b are such that a^2 = b.

If all digits are raised by one this is still valid. What are a and b?

Homework Equations


N/a

The Attempt at a Solution


Suppose a = a_0 \cdot 10^0 + a_1 \cdot 10^1

and b = b_0 \cdot 10^0 + b_1 \cdot 10^1 + b_2 \cdot 10^2 + b_3 \cdot 10^3

then it gets messy if I use a^2 = b and if I increase all digits by one.

I think should get two equations with 2 unknowns.
Start by looking at squares modulo 10 and think about conditions on ##a_0## and ##b_0##
 
dirk_mec1 said:

Homework Statement


Given are two integers a and b: a has 2 digits and b has 4 digits. All digits are smaller than 9. a and b are such that a^2 = b.

If all digits are raised by one this is still valid. What are a and b?

Homework Equations


N/a

The Attempt at a Solution


Suppose a = a_0 \cdot 10^0 + a_1 \cdot 10^1

and b = b_0 \cdot 10^0 + b_1 \cdot 10^1 + b_2 \cdot 10^2 + b_3 \cdot 10^3

then it gets messy if I use a^2 = b and if I increase all digits by one.

I think should get two equations with 2 unknowns.
I see no benefit in having separate variables for the digits.
The question wording is a little strange... in what sense 'valid'? I guess they mean that all the conditions for a and b are still true after elevating each digit. Because all digits are less than 9, you can very easily write an algebraic equation representing that the new a and b also satisfy the quadratic relationship. After that it's straightforward.
 
The nice thing with this problem is that you can narrow down your list of possible answers if you can never solve using relationships (although you should try to use them!). Because of the requirements, you know that a has to be between 32 and 87, with many of those numbers in there also not being possibilities. I wondered how long it would take to solve using guess and check (again, not the ideal method) and I guessed a on my first try! :)
Just to point out that there are always other approaches if you ever get really stuck on an exam or something
 
So solve:

a^2=b and <br /> (a+11)^2=(b+1111)

=> a =45 and b = 2025.
 
Last edited:
dirk_mec1 said:
So solve:

a^2=b and <br /> (a+11)^2=(b+1111)

=> a =45 and b = 2025.
That's what I get.
 
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a2=b

then:

(a+10+1)2=b+1111

this is the add one to each digit

a2+22a+121=b+1111

then rearrange onto one side

a2+22a-b-990=0

and as seen from the start

b=a2

so

a2+22a-a2-990=0

so

22a-990=0

so

22a=990

and

a=990/22
=45

then substitute this back into the original equation

452=b
b=2025

then to check:

562=3136

hope that helps
 
StanEvans said:
a2=b

then:

(a+10+1)2=b+1111

this is the add one to each digit

a2+22a+121=b+1111

then rearrange onto one side

a2+22a-b-990=0

and as seen from the start

b=a2

so

a2+22a-a2-990=0

so

22a-990=0

so

22a=990

and

a=990/22
=45

then substitute this back into the original equation

452=b
b=2025

then to check:

562=3136

hope that helps
This appears to be just an elaboration of post #6.
 
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haruspex said:
This appears to be just an elaboration of post #6.
yes sometimes it is easier explained
 
  • #11
StanEvans said:
yes sometimes it is easier explained
I don't think dirk_mec needs his own working explained to him.
 
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  • #12
haruspex said:
I don't think dirk_mec needs his own working explained to him.
What about others that are viewing the post?
 

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