# Square roots, integer 1

Square roots, integer

Doing square roots is just the opposite of squaring. This seems to force an open door. I hope to be able to explain you how important the basic knowledge of squares is to facilitate integer square roots.

Our Long, long ago my math teacher spoke about the check on an operation made.

Eg having done an integer sqrt ending on 1 the last digit of the answer should be a 1 or a 9. Similarly sqrt 4 last digit is 2 or 8, sqrt 6 is 4 or 6 etc. This brougt me to an idea to investigate the structure in the squares.

1² 1

2² 4

3² 9

4² 16

5² 25

6² 36

7² 49

8² 64

9² 81

10² 100

The left column speaks for itself. But have a look at the right one, and especially at the last digits of the squares. We see that the 1 did not change, neither did the 5 and the 6, and of course the 0. Furthermore the 4 got a 6, 2,and 8 got a 4 and 3 and 7 got a 9. In the right column 2,3,7 and 8 disappeared from the right column.

And there is more: 1+9=10, 2+8=10, 3+7=10 etc. So squares with identical last digits have a “mirror effect”.

This brought me to the question if on a bigger scale there were also such effects. As from my 10th year I know all the 2×2 multiplications by heart this investigation was not very difficult, as there are squares in too.

1² (0)1

49² 2401

51² 2601

99² 9801

Have a look at this table. Immediately you see the last digits of the squares all being 01. Besides 1+99=100, 49+51=100, 99-49=50, 51-1=50. Every mathematician can give you the explanation of this appearance.

√5184. The first digit of the answer must be 7 as 49<51<64. As the last digit of the radient is 4 the last digit of the answer has to be either 2 or 8.

Now there are more methods to find the 2nd digit of the answer. 70² + 2×2×70 = 5180, so it lies at hand that the final answer is 72.

One can also calculate:

√5184.

The first lower square is 7²=49, so the fist digit of the answer is 7. 51-49=2, to divide by 2 =1 and then to multiply with 10=10. Next is to append the half of the following digit, 8÷2=4. Add 10+4=14 and divide by 7=2, and the answer is complete: 72.

For bigger numbers knowledge of the squares till 100 is possibly not a must, however it is very skilful.

√66049.

As 625<660<676, the First digits of the answer are 25. Now subtract 660-625=35÷2=17.5

17.5×10 = 175 + append 4÷2=2. 175+2=177÷25=7 r 2. Answer so far 257.

2×10=20+(9÷2)= 24.5 – 7²/2= 0÷25=0, so answer 257,0. Suffice it to say that we write 257.

√46090521.

4489<4609<4624, so first digits 67.

4609-4489=120÷2=60×10 + half appended digit (0)= 600÷67=8r 64. Answer so far 678.

64×10 + half of appended next digit =642.5- 8²/2=610.5÷67=9 r 7.5. Answer so far 6789.

7.5×10 + half of appended 2= 76-8×9=4÷67=0 r 4. answer so far 6789,0

4×10 + half of appended 1 = 40,5-9²/2= 0÷67=0r 0 Answer so far 6789,00. We write 6789.

The bigger the number, the more challenging it will be.

√188077475041. Let’s write it as follows 188077 475041. He who is confident with the squares till 1000 will see that the three first digits of the answer are 433 ad 187489<188077<188356. Looking at f041 we’ll study the squares ending on 5041 and will soon see one of the candidates is 071 as this number squared is 5041. To get 75041 we have to get 70000. 70000÷2 =35000, and 35071²= 1229975041. And 75041= 1(32). So the basic number has to be either 1, 15, 17 or 31(32).

This is a good opportunity to introduce the quadrant/octant theme. It was Robert Fountain who draw my attention on this theme, which is a welcome enrichment for doing square roots.

We can find the quadrant by taking the first six digits of the radiant, and subtract the nearest lower square. Then multiply by 2 and divide the result by the first three digits of the answer.

188077- 187489= 588. 588×2=1176÷433=± 2,7. We have now the third quadrant. First quadrant 1-250, second 251-500, third quadrant 501-750, fourth quadrant 751-1000.

In this case as 2,7 is far over 2 the quadrant is 3: 501-750.

For finding the octant, which means more accuracy, we take the same difference, in this case 588 and multiply × 4=2352÷433=± 5,5 so the sixth octant. Octant 1 = 001-125, octant 2 = 126-250, octant 3 = 251-375, octant 4 = 376-500, octant 5 = 501-625, octant 6 = 626-750, octant 7 = 751-875, octant 8 = 876-1000.

In this case 626-750.

There are 8 numbers squared ending on 041, for of them OT, Odd Thousand: 071, 321, 679 and 929. For ending on 5041 there are 0071, 6321, 3679 and 8929. Out of these four there is one which is in the 6th octant, 3679, and now we can combine 433 and (3)679, to get 433679 as the final answer.

Check: the radiant is 18+80+77+47+50+41= 7+3+3+6+8=5(11). So the answer has to be 4(11).

43+36+79=10+3+2=4(11).

## 2 gedachten over “Square roots, integer 1”

Ik heb genoten van het lezen.

Dat is altijd leuk om te vernemen! W.B.