Finding a & b with Raised Digits

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The problem involves finding two integers, a and b, where a has two digits, b has four digits, and a^2 = b. When all digits of a and b are raised by one, the relationship still holds. The solution is derived by setting up equations based on the conditions and simplifying them. The correct values found are a = 45 and b = 2025, which satisfy both the original and modified conditions. The discussion emphasizes the importance of algebraic manipulation and checking solutions for accuracy.
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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
 
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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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