## Janet’s solution to Monday Math Madness #5

As promised, here is Janet's solution to Monday Math Madness #5.

Aside from 1 and 9, are there any perfect squares whose digits are all odd? Justify your answer.

No.

All perfect squares two digits and larger have at least one even digit.

Let’s see…

1. State as obvious that all squares of even numbers have an even digit.

2. Prove by brute force that all squares of single-digit odd numbers except those stated in the question have an even digit.

3. Need to show that all squares of multi-digit odd numbers have an even digit. Thankfully, there’s an even digit in the “10’s place”

Let D be a multi-digit odd number. Let’s represent this as a series of digits, so that

D = d

_{n}*10^{n}+ ….+ d_{1}*10 + d_{0 }

(where each d_{i}is in {0…9}, except d_{0}in {1,3,5,7,9} since D is odd)D

^{2}= (d_{n}*10^{n}+ … + d_{1}*10 + d_{0}) * (d_{n}*10^{n}+ … + d_{1}*10 + d_{0})We can multiply all these terms out with algebra, but we are only looking to prove that the digit in the 10’s place is even, so drop out all the terms multiplying by 10 to a higher power than one since they only contribute to the higher place value digits.

The only terms contributing to digits in the 10’s place are just these last two:

2 * 10d

_{1}d_{0}+ d_{0}^{2}d

_{0}is 1, 3, 5, 7, or 9, therefore d_{0}^{2}is 1, 9, 25, 49, or 81. The number “carried” into the 10’s place from this term is either 0 or even. Since odd + even is odd and even + even is even, the number being carried to the 10’s place will not change the oddness or evenness of that digit.The other term is 10 * 2d

_{1}d_{0}. Since 2 times anything is even, the last digit of 2d_{1}d_{0}is clearly even, which puts an even digit in the 10’s place. Even with a possible carried number from the lower order term, it will remain an even number.Therefore, the product of all two or more digit odd numbers contains an even digit in the tens place.

JonathanJune 12th, 2008 - 17:32

How did you figure this out, it seems to tuff for me :0