# Squaring a number near a base

Two weeks ago, I found YouTube videos discussing this problem and a similar one which is said to be a Math Olympiad problem, “20^18 – 1”.

The approach to solving the problems, as shown in both videos, are basically the same but I will discuss first our featured example.

2^20 can be expressed as 2^(10×2) which can be written as (2^10)^2.

In this form, (2^10)^2 – 20^2 is clearly a difference of two squares, which can be factored as (2^10 + 20) x (2^10 – 20).

In the video, the computation for 2^10 = 1024 was not shown.  But since almost all of us now are concerned with storage space either in computers or cell phones, it would be beneficial to know that 1 kb is not 1000 but 1024 bytes. While the prefix “k” means 1,000 in SI units, in computer jargon, 1 K = 1024, which is the value of 2^10.

So 1 Mb = 1k x 1kb, 1 Gb = 1k x 1 Mb, and 1 TB = 1k x 1 Gb.

We now have (1024 + 20) x (1024 – 20) or 1044 x 1004.

This was solved using the well known formula (x + a) (x + b) = x^2 + (a + b)x + ab with x = 1000.

I found this solution as a good example of using the difference of two squares so I used it in one of my IVMO 2022 Training sessions. But I used the Vedic  base multiplication to quickly determine the product 1044 x 1004 as (1044 + 4 | 44 x 4), which is 1048|176.

Later I realized the there is faster vedic math solution which can be done mentally.

We can directly square 2^10 or 1024 and then subtract 20^2 or 400. In squaring numbers above the base, we can use the Sutra “Whatever the excess, increase it further by that amount and and set-up the square of the excess”.

Here the excess from the base 1000 is 24 and we have a quick way of squaring 24. 24^2 = (25 – 1)^2 = 25^2 – 25 – 24 = 625 – 25 -24 =576.

Hence 1024^2 = (1024 + 24 | 24^2 = 1048 | 576

And 1, 048, 576 – 400 = 1, 048, 176

For numbers below the base, we will use the equivalent sutra: Whatever the deficiency, decrease it further by that amount and set-up the square of the deficiency.

Thus 8^2 = 8 -2 | 2^2 = 6 | 4

97^2 = 97 – 3 | 3^2 = 94 | 09

991^2 = 991 – 9 | 9^2 = 982 | 081

We will discuss similarly easy way to solve 2^18 – 1 next week.

Exercises: Find the value of the ff:

1. 12^2 =
2. 14^2 =
3. 17^2 =
4. 103^2 =
5. 106^2 =
6. 112^2 =
7. 1031^2 =
8. 98^2 =
9. 993^2 =
10. 989^2 =

Answers to last week’s exercises: Find the square root of the following:

1. 576 → 24
2. 1024 → 32
3. 1521 → 39
4. 1849 → 43
5. 2916 → 54
6. 4356 → 66
7. 5041 → 71
8. 6084 → 78
9. 7744 → 88
10. 8836 → 94