I’m preparing for a test and I keep getting stuck on questions like: without a calculator, decide if 529200 is a perfect square (and if not, get close to its square root). I know some basics: squares can only end in 0,1,4,5,6,9, and a square must have an even number of trailing zeros. I tried prime factorisation and the “all exponents even” idea, but doing the full factorisation feels too slow under time pressure. I also tried checking remainders mod 4 and 9, but I’m not sure how much that actually narrows things down. What are reliable, fast checks that work in practice? For example, are there useful rules about the last two digits (like 25 implying the root ends in 5), or quick bounds from estimating the square root that combine well with digit tests? Follow-up: is there a simple workflow you’d recommend-like first rule out by last digit, then use a quick sqrt estimate, then a modular check-or is there a better sequence I should learn?
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3 Responses
A handy workflow is to use quick sieves first, then a one-step square-root estimate. Think of it like airport security: a couple of fast gates catch most impostors before you do anything slow. My go-tos: (1) End-digit rules: a square can only end in 0,1,4,5,6,9; if it ends in 5 it must end in 25; if it ends in 0 it must end in 00; if it ends in 6 the tens digit must be odd; if it ends in 4 the tens digit must be even. Also, an even number of trailing zeros is required-strip zeros in pairs and test what’s left. (2) Mod tests that are quick in your head: digit sum mod 9 must be in {0,1,4,7}; last three digits mod 8 must be in {0,1,4}. These two together rule out a lot. (3) If it survives, bracket the root by nearby hundreds (700^2, 800^2, etc.), then refine with one smart try using (a+b)^2 = a^2 + 2ab + b^2 or do a tiny linear correction: if N is near a^2, then sqrt(N) ≈ a + (N − a^2)/(2a). It sounds fancy, but it’s just “how far off, spread over the slope 2a.”
Now your example, 529200. It has two trailing zeros, so if it were a square the root would end in 0; divide out 100 and the core 5292 would also have to be a perfect square. But look at its last two digits: 92 is impossible for a square (even-ending squares finish with 00, 04, 16, 24, 36, 44, 56, 64, 76, 84, or 96-no 92), so 529200 is not a square, ruled out in seconds. To get close to the root anyway, bracket: 700^2 = 490000 and 800^2 = 640000, so it’s between 700 and 800. Try 728: 728^2 = (700+28)^2 = 490000 + 39200 + 784 = 529984, a bit high; 727^2 = 528529, a bit low; so sqrt(529200) is about 727.5. With the linear tweak: 728 − (529984 − 529200)/(2·728) = 728 − 784/1456 ≈ 727.46. Quick, tidy, done.
A fast, reliable workflow I use is: (1) strip easy powers of 100 (and note that a square must have an even number of trailing zeros; if it ends in 00 then the root ends in 0). (2) Run quick filters: last digit must be 0,1,4,5,6,9; if it ends in 5 it must end in 25; if it’s even but not divisible by 4, it’s not a square. Modular checks that prune a lot quickly: mod 9 must be in {0,1,4,7} and mod 16 must be in {0,1,4,9}. If you want a stronger digit test, the last two digits of a square must be one of 00,01,04,09,16,21,24,25,29,36,41,44,49,56,61,64,69,76,81,84,89,96; anything else is out. (3) If it passes filters, bracket the root: find A with A^2 ≤ N < (A+1)^2 using nearby round numbers (e.g., compare with 700^2, 720^2, 730^2, …). A quick refinement is sqrt(N) ≈ A + (N − A^2)/(2A), then just test the one or two nearest integers by squaring. Example (your 529200): strip 100 → 529200 = 5292 × 100, so any square root must be 10 times an integer, and we only need to check whether 5292 is a square. A last-two-digits check already rules it out: 92 is not in the allowed list, so 5292 is not a square, hence 529200 is not a perfect square. To get close to the root, note 72^2 = 5184 and 73^2 = 5329, so sqrt(5292) is between 72 and 73; therefore sqrt(529200) is between 720 and 730. A quick linear tweak from 720: 720^2 = 518400, difference 10800, and 2·720 = 1440, so add about 10800/1440 ≈ 7.5 to get ≈ 727.5. Indeed, 727^2 = 528529 and 728^2 = 529984, so 529200 sits between them; the integer square root is 727, so it’s not a perfect square.
When I’m speed-detecting squares, I use a little “square-sniff” routine: first, the last-digit gate (kick out 2, 3, 7, 8). If it gets through, do the 00 peek: whenever a perfect square ends with 00, the hundreds digit is always 4 or 6-squaring numbers ending in 0 builds chunks like 400, 1600, 3600, 6400, and beyond that the hundreds place just keeps flipping between 4 and 6-so any 00-number with a different hundreds digit can’t be a square. Then I corral the number between friendly squares and walk linearly across the gap using the almost-constant step size near n, which is about 2n+1. Worked example: 529200 ends with 00, but its hundreds digit is 2, so it fails the 00-hundreds test-definitely not a square. For the root, bracket it: 700^2=490000 and 750^2=562500, a gap of 72500 across 50 steps, so each step is about 1450. Since 529200−490000=39200, that’s roughly 39200/1450≈27 steps past 700, so √529200≈727; indeed 727^2=528529 and 728^2=529984, putting the true root around 727.7. My quick workflow: last digit, 00-hundreds check if relevant, then bracket-and-interpolate-swift, tidy, and delightfully square-ish.