Theory
Squaring numbers that end in 5
What is 65 squared? Most people reach for pen and paper. But there is a beautifully simple rule for squaring any number ending in 5, and it comes from the sutra Ekadhikena Purvena, 'by one more than the previous one'.
This is the second technique in the unit. Like Nikhilam, it is a systematic shortcut you can learn, check against ordinary arithmetic, and code. As before, we treat it purely as the calculation method it describes, from Bharati Krishna Tirtha's 1965 book, and work verified examples.
Theory
The rule
To square a number ending in 5:
1. Look at the part before the 5 (the sutra calls it the 'previous'). For 65, that part is 6.
2. Multiply it by one more than itself ('by one more than the previous one'). One more than 6 is 7, so 6 times 7 is 42.
3. Append 25 to the result.
So 65 squared is 42 | 25 = 4225. The last two digits are always 25 (because a number ending in 5, squared, always ends in 25); all the work is the single small multiplication of the front part by the next integer up.
Theory
More examples
The rule is the same every time.
- 25 squared: front part 2, times one more (3), is 2 times 3 = 6; append 25 -> 625.
- 35 squared: 3 times 4 = 12; append 25 -> 1225.
- 65 squared: 6 times 7 = 42; append 25 -> 4225.
Each checks out against ordinary squaring. Once you see it, squaring 15, 25, 35, all the way up becomes something you can do instantly in your head: multiply the front by the next number, stick 25 on the end.
Practical
Ekadhikena Purvena in Python (verified)
def square_ending_in_5(n):
# n must end in 5, e.g. 25, 35, 65
front = n // 10 # the part before the 5 (65 -> 6)
left = front * (front + 1) # 'by one more than the previous': 6 * 7 = 42
return left * 100 + 25 # append 25
print(square_ending_in_5(25)) # 625
print(square_ending_in_5(35)) # 1225
print(square_ending_in_5(65)) # 4225This example runs in Gri-Learn on the web, where you can edit it and see the output.
Quiz
Using Ekadhikena Purvena, what is 35 squared?
- 1215, appending 15
- 1225, because 3 times (3+1) = 12, then append 25
- 925, because 3 times 3 = 9, append 25
- 1250, because 35 times 35 rounds to 1250
Show the answer
1225, because 3 times (3+1) = 12, then append 25
Take the part before the 5, which is 3, and multiply by one more than itself: 3 times 4 = 12. Then append 25 to get 12 | 25 = 1225, which is indeed 35 squared. Option A wrongly appends 15; the last two digits of the square of a number ending in 5 are always 25. Option C uses 3 times 3 (the number itself) instead of 3 times 4 (one MORE than itself), missing the whole point of 'by one more than the previous'. Option D is a vague rounding guess with no basis. The rule is exact: front times (front + 1), then 25.
Think first
Why does 'front times one more, then 25' work?
Why does multiplying the front by the next integer and appending 25 give the exact square? Then tap.
Show the answer
Because it is the algebra of squaring (10a + 5), where a is the front part. Any number ending in 5 can be written as 10a + 5 (for 65, a is 6, giving 60 + 5). Square it: (10a + 5) squared = 100a squared + 2(10a)(5) + 25 = 100a squared + 100a + 25 = 100 times (a squared + a) + 25 = 100 times a(a + 1) + 25. Read that result: the term a(a + 1) is 'the front times one more than itself' (for a = 6, that is 6 times 7 = 42), and multiplying it by 100 shifts it into the hundreds, while the + 25 fixes the last two digits as 25. So 65 squared is 100 times 42 plus 25, which is 4200 + 25 = 4225, exactly the '42 then 25' the rule produces. The appended 25 is not a coincidence; it falls straight out of the 5 squared term. So the sutra is a memorable packaging of a clean algebraic identity, which is why it always works for numbers ending in 5, and why the front is multiplied by 'one more than itself' rather than by itself. Simple identity, elegant shortcut.
Summary
Key takeaways
- Ekadhikena Purvena ('by one more than the previous one') squares any number ending in 5.
- Take the part before the 5, multiply it by one more than itself, then append 25.
- 65 squared: 6 times 7 = 42, append 25 -> 4225; 25 squared -> 625; 35 squared -> 1225.
- The last two digits are always 25, so only the small front multiplication varies.
- It works because (10a + 5) squared = 100 times a(a + 1) + 25, plain algebra.
- It is a technique from the 1965 book by Bharati Krishna Tirtha.
- Memory hook: front times the next number, then stick 25 on the end.