Here is the method I used to solve the original poster's problem in about three seconds:
Square root (5329):
Steps 1a-d estimate the answer, to one significant figure, with the correct order of magnitude:
1a) Start with an estimated answer of 1.
1b) Find the order of magnitude of the answer. Make a copy of the argument (5329). Shift the decimal of the copy of the argument by two places at a time, to get a number between 1 and 100. Shift the decimal of the estimated answer by one place at a time, in the opposite direction. For example, 53.29 and 10.
1c) Assume that the student knows the squares of the integers between 0 and 10 by heart. What is the closest such square? For example, 49.
1d) Multiply the estimated answer by the square root of the closest square. For example, square root (49) = 7, so the new estimated answer is 70.
Step 2 checks the estimate:
2) Square the current estimated answer. For example, 70 * 70 = 4900.
Steps 3a-d refine the estimate using uses (a+b)^2 = a^2 + 2*a*b + b^2. This is a special case of Newton's method:
3a) Subtract the value found in step 2 from the original argument, to yield a remainder. For example, 5329 - 4900 = 429. Keep track of the sign.
3b) Make a copy of the estimated answer. Multiply the copy by 2, to yield a double estimate. For example, 70 * 2 = 140.
3c) Divide the remainder by the double estimate, to yield an adjustment. Round off as convenient. Keep track of the sign. For example, 429 / 140 ~ 3.
3d) Add the adjustment to the estimate (not the double estimate). We kept track of the sign, so that we could correctly choose whether to add or subtract. For example, 70 + 3 = 73. For subsequent calculations, this value replaces estimate.
Repeat steps 2 - 3 until the answer is close enough. For example, 4900 + 420 + 3*3 = 5329, so the square root of 5329 is 73.
Some advantages of this method:
I can do it in my head.
It only requires keeping track of a few numbers.
It quickly produces an estimated answer.
It is a special case of Newton's method.
Each time through the loop, the student can round off the adjustment to a convenient decimal or fraction.
Every iteration, the student checks their work.
Three worked examples:
Argu- Estimated Found Double Adjust-
Step ment Answer So Far Remainder Estimate ment Notes
1a 5329 1
1b 10 53.29
1c 7*7=49
1d 70
2 5329 70 4900
3a 5329 70 4900 429
3b 5329 70 4900 429 140
3c 5329 70 4900 429 140 3
2 5329 73 5329 73*73=4900+420+9
Done!
Argu- Estimated Found Double Adjust-
Step ment Answer So Far Remainder Estimate ment Notes
1a 739 1
1b 10 7.39
1c 3*3= 9
1d 30
2 739 30 900
3 739 30 900 -161 60 -3
2-3 739 27 729 10 54 0.2
2-3 739 27.2 739.84 -0.84 54.4 -0.015
2-3 739 27.185 739.024225 -0.024225 54.37 -0.0005
2-3 739 27.1845 738.99704025 54.369 0.00005
2 739 27.18455 738.9997587025
...Stop when satisfied...
Argu- Estimated Found Double Adjust-
Step ment Answer So Far Remainder Estimate ment Notes
1a 4285 1
1b 10 42.85
1c 6*6=36
1d 60
2-3 4285 60 3600 685 120 5
2-3 4285 65 4225 60 130 0.4
2-3 4285 65.4 4277.16 7.84 130.8 0.06
2-3 4285 65.46 4285.0116 -0.0116 130.92 -0.0001
2-3 4285 65.4599 4284.99850801 130.9198 0.00001
2 4285 65.45991 4284.9998171081
...Stop when satisfied...