Bmo 1 1996 q1

Is the solution just 3600,2500 and 0036,0025.
Reasoning: Since there needs be a difference of 1s and we know that M,N are squares. a^2-b^2=(a+b)(a-b) and the differences must be one of {1100,1010,1001,0110,0101,0011}, and u can show quite quickly that for all other than 1100 and 0011, there exists no solutions. leaving the answers 3600,2500 and 0036,0025. To show that they don't satify, break down the difference into prime factorisation and there'll be 2C(number of factors) possible combinations, and u can eliminant most since odd(O)+/-even(E)=O, O+/-O=E, E+/-E=E. Then just add the numbers (a+b),(a-b) together and half it to get 'a', then square 'a' and see whether it has two of the same numbers in the places of the '00', for most cases, this doesn't match, so its not a possible pairing, so eliminate it, then we get left with answer pairs of 3600,2500 and 0036,0025. If 4-digit integers counts for something that begins with 0, then keep 0036,0025; else there's only one pair.

I think you've missed some solutions.

Its also worth noting that 0036, 0025 are not a 4 digit numbers. The leading digit has to be nonzero which gives a range of values of the two numbers etc. But as dfranklin notes you obv need to revisit how you justify there are no other solutions near the start. A difference of 0101 should almost be trivial (consecutive squares).

1024 cannot be one, so the pair fails the criteria, u need two digits to be the same and two different as the criteria.

The factors of 1001 are 1, 7, 11,13, 77, 91, 143 and 1001. Dfranklin is referring to the 13,77 factor pair.

I noted at the end of a number starting with 0 doesn't count as 4 digits persay, we ignore the 0036,0025. A difference of 0101 is possible but doesn't satisfy the criteria's, 101=1x101. So it's a+b and a-b would be 51 and 50. Since they are different, in that one is odd and another is even, it breaks one of the supposed criteria's by me, so it's not possible. The only other thing I had not checked was the cases where u would be adding 1 to a 9, then potentially they'll be more solutions, but I doubted that could satisfy so didn't check them.

Dont understand your logic
51^2-50^2 = 2601 - 2500 = 0101
so its a solution? If not, explain why in terms of how the question is asked, rather than extra logic youve added.

Please read the question first 😭 2601 fails automatically

Why? Its a square with two equal digits and two digits differ by 1 from 2500

I've attached the question, please read the bold part saying that there must be 2 equal digits and differing digits.

Ill drop out. 2601 is a square 51^2 and 2500 is a square 50^2 and there are two equal digits
2@0@
and two digits that differ by 1
@(5+1)@(0+1)

Thxs for trying, but the main issue with this pair is that not both 4 digit numbers satisfy all the conditions of the question. Yes, 2500 satisfy, but 2601 does not.

Thxs for trying, but the main issue with this pair is that not both 4 digit numbers satisfy all the conditions of the question. Yes, 2500 satisfy, but 2601 does not.

Additionally, if it were 2500, you would have to add 1100 because of the question saying differing only in the places of differing numbers, so if a number was xx12, you have to add 1s where the 12 is not any combination you desire.

Moreover, when the digits differ, the digit in
M is exactly one greater than the corresponding digit in N.

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From the question ^