Is he wrong?

Number Theory Level 2

Hamza multiplied two 2-digit number and then he wrote the equation on the blackboard (For example 13 × 10 = 130 13 \times 10 = 130 ). He then changed the digits with letters. Similar digits where changed by the same letter and no two digits were changed by the same letter.

If he wrote A B × A B = C C D D AB \times AB = CCDD

Is Hamza wrong?

Yes No

Mohd. Hamza by Mohd. Hamza , 18, India

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2 solutions

Mohd. Hamza
Nov 27, 2018

The LHS is not divisible by 11 but the RHS is. So Hamza made a mistake.

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U can write A B AB as 10 A + B 10A + B

A B × A B = ( 10 A + B ) 2 = 100 A 2 + 20 A B + B 2 = C C D D = 1000 C + 100 C + 10 D + D . AB\times AB = (10A + B) ^2 = 100A^2 + 20AB + B^2 = CCDD= 1000C + 100C + 10D + D.

We can clearly see that, for this to happen we at least need 100 A 2 = 1000 C ⇒ A 2 = 10 C 100A^2 = 1000C \Rightarrow A^2 = 10C and B 2 = 10 D + D . B^2 = 10D + D.

Now, one one thing left 20 A B = 20 10 C 10 D + D ≠ 100 C 20AB = 20\sqrt {10C} \sqrt {10D + D } \neq 100C

This cannot take place since you can't take out 100 out as for that you need D D to be at least 10 ,which contradicts that D D is a single digit positive integer.

Hence, Hamza (I see that he himself is the writer of this problem) is W R O N G . \boxed{WRONG}.

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