Question

find 256 square root by repeated subraction​

Answer 1

Answer:

under root 256=16

Step-by-step explanation:

Step 1: 256 - 1 = 255

/* Subtracting of first odd number */

Step 2: 255 - 3 = 252

/* Subtracting of second odd number */

Step 3: 252 - 5 = 247

/* Subtracting of Third odd number */

Step 4: 247 - 7 = 240

Step 5: 240 - 9 = 231

Step 6: 231 - 11 = 220

Step 7: 220 - 13 = 207

Step 8: 207 - 15 = 192

Step 9: 192 - 17 = 175

Step 10: 175 - 19 = 156

Step 11: 156 - 21= 135

Step 12: 135 - 23 = 112

Step 13: 112 - 25 = 87

Step 14: 87 - 27 = 60

Step 15: 60 - 29 = 31

Step 16: 31 - 31 = 0

From 256,we have subtracted 16 successive odd numbers starting from 1 and obtained zero (0) at 16 th step.

So under root 256=16

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Answer 2

$\huge\red{\mathfrak{hello!!}}$

Answer:

Subtracting of first odd numbers

Step 1: 256 - 1 = 255

Step 2: 255 - 3 = 252

Step 3: 252 - 5 = 247

Step 4: 247 - 7 = 240

Step 5: 240 - 9 = 231

Step 6: 231 - 11 = 220

Step 7: 220 - 13 = 207

Step 8: 207 - 15 = 192

Step 9: 192 - 17 = 175

Step 10: 175 - 19 = 156

Step 11: 156 - 21= 135

Step 12: 135 - 23 = 112

Step 13: 112 - 25 = 87

Step 14: 87 - 27 = 60

Step 15: 60 - 29 = 31

Step 16: 31 - 31 = 0

From 256,we have subtracted 16 successive odd numbers starting from 1 and obtained zero (0) at 16 th step.

√256 = 16

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