# The First by the First and the Last by the Last

11^{2} + 22^{2} + 33^{2} + 44^{2} + 55^{2} = ?

a) 4505

b) 5566

c) 5808

d) 6655

e) 7105

This question was given in the Juniors and older age groups in the 1^{st} Math2Shine International Vedic Mathematics Competition which was held last July 22, 2023.

A variation of this question, 22^{2} + 33^{2} + 44^{2} = ?, was also given in the Beginners and Primary categories.

This is a very easy question, but it would involve lengthy calculations using the traditional methods.

Most people would square all 5 numbers and then add the squares:

11^{2} = 121

22^{2} = 484

33^{2} = 1089

44^{2 }= 1936

55^{2} = 3025

121 + 484 + 1089 + 1936 + 3025 = 6655

Many would find that computing 11^{2} + 22^{2} + 44^{2} first would save time, since the expression can be converted to 11^{2} +(2 x 11)^{2} + (4 x 11)^{2} or (1 + 4 + 16) x 11^{2} which reduces to a simple (21 x 121= 2541) multiplication.

In the Beginners and Primary version of the question, getting the value of 22^{2} + 44^{2} is easier. (2 x 11)^{2} + (4 x 11)^{2} = 20 x 121 = 2420.

The Vedic Math solution of applying the Sutra, **the First by the First and the Last by the Last,** would not give us the exact answers to the problems but would give us the best among the choices offered.

We begin by squaring the last digits of the numbers **(the Last by the Last)** and adding as we go along:

1^{2} + 2^{2} = 1 + 4 = 5

5 + 3^{2} = 14

14 + 4^{2} = 30

And 30 + 5^{2} = 55.

Therefore, the answer must end in 5 and we can eliminate b) 5566 and c) 5808. We can also eliminate choice a) 4505 because the first two digits of the answer must be greater than 55.

We are thus, left with two choices d) 6655 and e) 7025.

The next step is to estimate the value of 15^{2} + 25^{2} + 35^{2} + 45^{2} + 55^{2} because the value we get here is very much more than the answer to our problem. Applying “**by one more than the one before**” to the first digits of the numbers to be squared **(the First by the First)** and adding them as we go along, we have:

1 x 2 = 2

2 + (2 x 3) = 8

8 + (3 x 4) =20

20 + (4 x 5) = 40

40 + (5 x 6) = 70

We cannot have e) 7025 as the answer. So, what remains is d) 6655.

Can you apply this method in solving the version for the Beginners and Primary group?