A number theory problem by Alf-archie Sangkula
Number Theory
Level
pending
Find the largest even four-digit perfect square whose units digit exceeds the tens digit by 1 and whose thousands digit exceeds the hundreds digit by 1.
4356
4365
6543
3456
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1 solution
The smallest four-digit perfect square is 32^2 = 1024 while the largest is 99^2 = 9801. Hence, there are 99 – 32 + 1 = 68 cases to check in an exhaustive search. We can strike out about half of them since we are only looking for even perfect squares. So, if you are really desperate, exhaustive search may not be too bad an idea (and I would recommend starting from 98^2 and working downwards since we are looking for the biggest one).
Let’s work out a better way. Let the perfect square be n^2. It is quite clear that n must be a 2-digit number, so let n = 10a + b, where a and b each represent a digit from 0 to 9 (inclusive). Then,
n^2 = (10a + b)^2 = \boldsymbol{100a^2 + 20ab + b^2}.
Let’s worry about the condition concerning the thousands & hundreds digit later: conditions concerning the tens & ones digit are generally a lot easier to tackle. What can the expression above tell us about the ones and tens digit?
First, only the b^2 term contributes to the ones digit, hence the choice of b completely determines the ones digit. Since n^2 must be even, we have 5 choices for b: 0, 2, 4, 6 and 8. However, the ones digit must be greater than 0 (why?), hence b cannot be 0.
Next, the 20ab and b^2 terms both contribute to the tens digit. However, notice that the 20ab term always contributes an even number to the tens digit. Hence, in order for the tens digit to be one less than the ones digit, it must be odd and so the b^2 term must contribute an odd number to the tens digit. This rules out b = 2 and b = 8, so b is either 4 or 6.
We have significantly reduced the number of cases! Now let’s go through each of them.
Case 1: b = 4. Then,
n^2 = 100a^2 + 80a + 16.
In order for the tens digit to be 5, a must be 3 or 8. However, 34^2 = 1156 and 84^2 = 7056 both do not satisfy the condition concerning the thousands and hundreds digit.
Case 2: b = 6. Then,
n^2 = 100a^2 + 120a + 36.
In order for the tens digit to be 5, a must be 1 or 6. 16^2 = 256 is not even 4 digits, while 66^2 = 4356 satisfies all the given conditions.
Hence, the answer is 4356. (In fact, it is the only 4-digit even perfect square to satisfy the given conditions.)