# Number problem

1. Mar 1, 2016

### dirk_mec1

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
N/a

3. 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.

2. Mar 1, 2016

### Epiclightning

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.

3. Mar 1, 2016

### PeroK

Start by looking at squares modulo 10 and think about conditions on $a_0$ and $b_0$

4. Mar 1, 2016

### haruspex

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.

5. Mar 1, 2016

### mattbeatlefreak

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

6. Mar 1, 2016

### dirk_mec1

So solve:

$a^2=b$ and $(a+11)^2=(b+1111)$

=> a =45 and b = 2025.

Last edited: Mar 1, 2016
7. Mar 1, 2016

### haruspex

That's what I get.

8. Mar 6, 2016

### StanEvans

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

9. Mar 6, 2016

### haruspex

This appears to be just an elaboration of post #6.

10. Mar 7, 2016

### StanEvans

yes sometimes it is easier explained

11. Mar 7, 2016

### haruspex

I don't think dirk_mec needs his own working explained to him.

12. Mar 12, 2016

### StanEvans

What about others that are viewing the post?