In competitions such as the JMCWC and MSO World Championship, some questions require calculation of square roots whose answer is an integer (a whole number).
For example, the square root of 900 is 30, because 30² = 900.
For example, the square root of 987 is 31.416556…, but that is not an integer. So it doesn’t make sense to find the “exact square root” of 987.
As we know that the answer must be an integer, the question is easier, because we can look at both ends of the question separately (the left end and the right end) to easily find different information.
Squares up to 99²
In order to efficiently find exact square roots of numbers of 6 or more digits, it is important to know the 2-digit square numbers from memory. Then it is quite easy to find the exact square roots of most 6-, 7- and 8-digit numbers. It is also practical to find exact square roots of even larger numbers—as you will see!
Here are the square numbers from 0² to 99². What patterns can you see in the last two digits?
| 00 ↘ 25 | 25 ↗ 50 | 50 ↘ 75 | 75 ↗ 99 |
|---|---|---|---|
| 00² = (0)0 | — | 50² = 2500 | — |
| (0)1² = (0)1 | 49² = 2401 | 51² = 2601 | 99² = 9801 |
| (0)2² = (0)4 | 48² = 2304 | 52² = 2704 | 98² = 9604 |
| (0)3² = (0)9 | 47² = 2209 | 53² = 2809 | 97² = 9409 |
| (0)4² = 16 | 46² = 2116 | 54² = 2916 | 96² = 9216 |
| (0)5² = 25 | 45² = 2025 | 55² = 3025 | 95² = 9025 |
| (0)6² = 36 | 44² = 1936 | 56² = 3136 | 94² = 8836 |
| (0)7² = 49 | 43² = 1849 | 57² = 3249 | 93² = 8649 |
| (0)8² = 64 | 42² = 1764 | 58² = 3364 | 92² = 8464 |
| (0)9² = 81 | 41² = 1681 | 59² = 3481 | 91² = 8281 |
| 10² = 100 | 40² = 1600 | 60² = 3600 | 90² = 8100 |
| 11² = 121 | 39² = 1521 | 61² = 3721 | 89² = 7921 |
| 12² = 144 | 38² = 1444 | 62² = 3844 | 88² = 7744 |
| 13² = 169 | 37² = 1369 | 63² = 3969 | 87² = 7569 |
| 14² = 196 | 36² = 1296 | 64² = 4096 | 86² = 7396 |
| 15² = 225 | 35² = 1225 | 65² = 4225 | 85² = 7225 |
| 16² = 256 | 34² = 1156 | 66² = 4356 | 84² = 7056 |
| 17² = 289 | 33² = 1089 | 67² = 4489 | 83² = 6889 |
| 18² = 324 | 32² = 1024 | 68² = 4624 | 82² = 6724 |
| 19² = 361 | 31² = 961 | 69² = 4761 | 81² = 6561 |
| 20² = 400 | 30² = 900 | 70² = 4900 | 80² = 6400 |
| 21² = 441 | 29² = 841 | 71² = 5041 | 79² = 6241 |
| 22² = 484 | 28² = 784 | 72² = 5184 | 78² = 6084 |
| 23² = 529 | 27² = 729 | 73² = 5329 | 77² = 5929 |
| 24² = 576 | 26² = 676 | 74² = 5476 | 76² = 5776 |
| 25² = 625 | — | 75² = 5625 | — |
Useful Patterns within the Squares up to 99²
Notice in the table that in any row, the last two digits are always the same—for example, 17, 33, 67 and 83. In general, if 0 < x < 25, then the following four numbers have the same last two digits:
- x²
- (50 – x)²
- (50 + x)²
- (100 – x)²
(There is a proof of this in the appendix.)
Furthermore, notice that each row has a different two-digit ending—except for multiples of 10, whose squares end in 00. This is useful when we want to find an exact square root, because the last two digits tells us which row of the table has the possible endings for our square root.
Notice also that the only possible two-digit endings are:
- 00 and 25
- 01, 21, 41, 61 and 81
- 04, 24, 44, 64 and 84
- 09, 29, 49, 69 and 89
- 16, 36, 56, 76 and 96
Therefore (for example) 27811346 cannot be a perfect square, because no perfect square can end in __46.
How to Calculate Exact Square Roots of Large Numbers
When you find the square root of a number with 7 or 8 digits, the square root will have at 4 digits. You can use the last two digits of the numbers to suggest what the last two digits of the square root could be. But you find the first digits, the most suitable method is to use the approximation here: estimating square roots. (This is enough for square roots of numbers up to about 12 digits. For most larger numbers, use this more advanced square roots algorithm gives better approximations).
Example: find the exact square root of 20857489
Step 1: starting from the right, split groups of 2 digits until a 3- or 4-digit number remains: From 20857489 we have 2085.
Use the table above to find the first two digits by finding the square beneath this. For 2085. this is after 2025 = 45² (but before the next square, 2116).
So the first digits must be 45. Two blocks of two digits follow, so there must be two digits remaining in the answer: 45 _ _.
Step 2: using the method in the first link above (not explained here), estimate (2085 – 2025) ÷ (2 × 45²) = 0.666666…, so the answer is approximately 4566.6666…. In fact, such precision is unnecessary: we could have used 0.66 or 0.67 or even 0.70 as the approximation instead, and concluded that the answer is close to 4566, 4567 or 4570.
Step 3: consider the last two digits of the number. In 20857489, it’s 89. Which squares end in __89?
Use the table above (from memory) to find the possible endings: 89 → 17, 33, 67 or 83.
However, we know that the true answer is close to e.g. 4570. Therefore the true answer is surely 4567.
Example: find the exact square root of 759405930721
Step 1: starting from the right, split groups of 2 digits until a 3- or 4-digit number remains: From 759405930721 we have 7594.
Use the table above to find the first two digits by finding the square beneath this. For 7594. this is after 7569 = 87² (but before the next square, 7744).
So the first digits must be 87. Four blocks of two digits follow, so there must be four digits remaining in the answer: 87 _ _ _ _.
Step 2: using the method in the first link above (not explained here), estimate (7594 – 7569) ÷ (2 × 87²) = 0.14367816…, so the answer is approximately 871436.7816…. In fact, such precision is unnecessary: we could have used 0.143 or 0.144 as the approximation instead, and concluded that the answer is close to 871430 or 871440.
Step 3: consider the last two digits of the number. In 759405930721, it’s 21. Which squares end in __21?
Use the table above (from memory) to find the possible endings: 21 → 11, 39, 61 or 89.
However, we know that the true answer is close to e.g. 871436. Therefore the true answer is surely 871439.
Difficult Ending-Digits
Squares ending in 00: to find the square root of a number ending in 00, ignore the 00s at the end, and place a 0 at the end of your answer for each 00 that was ignored. Example: square root of 23716000000 = (square root of 23716) × 1000 = 154000 (easy).
Squares ending in 25: between 00–99, there are ten numbers (not only four numbers) whose squares end in 25: these are 05, 15, …, 85 and 95. This can make it more difficult to determine the correct last digits! However, you can use the following rule to reduce the possibilities to maximum four:
- square ends in 025 → square root ends in 05, 45, 55 or 95
- square ends in 225 → square root ends in 15, 35, 65 or 85
- square ends in 625 → square foor ends in 25 or 75
Squares ending in 76: the square root could end in 24, 26, 74 or 76. Since 24 and 26 are so close together (likewise 74 and 76), how can we determine which one in the pair it is? Look at the hundreds digit (i.e., 3rd from the end). The outside (24 and 76) possibilities have an odd hundreds digit; the inside (26 and 74) have an even hundreds digit (proof in the appendix).
- outside: squares ending in 24 include 576, 15376 and 50176. Their hundreds digits are all odd.
- inside: squares ending in 26 include 676, 15876 and 51076 . Their hundreds digits are all even.
- inside: squares ending in 74 include 5476, 30276 and 75076. Their hundreds digits are all even.
- outside: squares ending in 76 include 5776, 30976 and 76176. Their hundreds digits are all odd.
When I discovered this rule, I called it outside–odd; inside–even or OO–IE to help me remember.
Other difficult endings have similar rules but they are more complex. For example, the rules for squares ending in 04 and 84 are opposite to these (outside even).
Proofs of Previous Statements
You can skip these proofs if you are not currently curious about the mathematics behind the method.
Proof that every square in the row has the same last two digits
Take any number, and write it in the form 50n ± x, where n and x are integers, and 0 < x ≤ 25. (For example 337 = 50 × 7 – 13)
Square it, expand the brackets and then attempt to isolate any multiples of 100:
(50n ± x)² = 2500n² ± 100nx + x² = 100(25n² ± nx) + x²
The last two digits of this are unaffected by n, and unaffected by whether the ± is a + or a –. Therefore (50n ± x)² always has the same last two digits as x². [Each row of the table shows square of 0 + x, 50 – x, 50 + x and 100 – x].
Proof of “outside–odd; inside–even”
Outside: Any number ending in 24 or 76 can be written as 100n ± 24. Its square is (100n ± 24)² = 100²n² ± 2400n + 24² = 200(50n² ± 12n) + 576
Since the hundreds-digit of 200(50n² ± 12n) is even, and the hundreds digit of 576 is odd, their sum must have an odd hundreds digit.
Inside: Any number ending in 26 or 74 can be written as 100n ± 26. Its square is (100n ± 26)² = 100²n² ± 2600n + 26² = 200(50n² ± 13n) + 676
Since the hundreds-digit of 200(50n² ± 13n) is even, and the hundreds digit of 676 is even, their sum must have an even hundreds digit.
Further Reading
You might be interested in:
- Information about international mental calculation competitions.
- A similar method for exact cube roots.
- More advanced methods for mental calculation including methods for square roots.
To get in touch with me (Daniel Timms) about mental calculation training, coaching, or anything on this site, you can contact me here.