We had discussed this method in our previous articles for 2-digit numbers and today we shall explain the same method for 3-digit numbers.

Example:

Here we can see that right digits sum is 100 i.e.(92 + 8) and left side digits are same. Here we can now apply the same method, which we discussed earlier for 2-digit number. But this time we must expect to have four figures on the right-hand side.

Example:

Here 95 + 05 = 100 and left side digits are same i.e. '7'. Hence it qualifies for this case.

In calculation, we shall multiple the last two digits and the left digit i.e. '7; with its next number '8'. So the calculation is:

795 x 705 = 7 x 8 | 95 x 05

= 56 | 0475

=

Example:

Here 66 + 34 = 100 and left side digit is 8 and its next number is 9. So the calculation is:

848 x 852 = 8 x 9 | 66 x 34 (Note: For 66 x 34, we shall discuss in our upcoming articles)

= 72 | 2244

=

Example:

Here we can see that right digits sum is 10 i.e.(9 + 1) and left side digits are same i.e. 24. So we can now apply the same method as described above.

241 x 249 = 24 x 25 | 1 x 9

= 0600 | 09

=

444 x 446 = 44 x 45 | 4 x 6

= 1980 | 24

= 198025

147 x 143 = 14 x 15 | 7 x 3

= 0210 | 21

= 21021

For example:

93 × 39 doesn't seems to comes under this method. But if we convert 93 to (3 × 31), then 31 × 39 comes under this method. So the calculation is

93 x 39 = 3 x 31 x 39

= 3 x (31 x 39)

31 × 39 = 1209 (by same method discussed above i.e. 31 x 39 = (3 x 4) | (1 x 9) = 12 | 09 = 1209)

93 × 39 = 3 x 1209

=

The point to notice in the above example is that '39' needs '31' to qualify for method, and then we found that 93 is '3 × 31'.

In 35 x 37, both left hand numbers are 3, but 5 + 7 = 12 which is greater than 10. We can solve this in

35 x 37 = 35 x (35 + 2)

= (35 x 35) + (35 x 2)

= 1225 + 70 (Take 35 x 35 = 3 x 4 | 5 x 5 = 12 | 25 = 1225)

=

So, we rewrite it as:

35 x 37 = (33 + 2) x 37

= (33 x 37) + (2 x 37)

= 1221 + 74 (Take 33 x 37 = (3 x 4) | (3 x 7) = 12 | 21 = 1221)

= 1295

One more example:

= (114 x 116) + (114 x 1)

= 13224 + 114 (Take 114 x 116 = (11 x 12) | (4 x 6) = 132 | 24 = 13224)

=

OR

114 x 117 = (111 + 3) x 117 (Rest of the procedure is same)

Now, take one more example where first digit remains same but addition of last digits is less than 10.

68 x 61 = 68 x (62 - 1)

= (68 x 62) - (68 x 1) (Rest of the procedure is same)

OR

68 x 61 = (69 - 1) x 61

= (69 x 61) - (61 x 1) (Rest of the procedure is same)

Now let us solve some case where

Here we can see that left hand digits sum is 10 i.e.(4 + 6) and right hand digits are same i.e. 7 . So we can now apply the following method.

Let us take some more examples

34 x 74 = (3 x 7) + 4 | 4 x 4

= 21 + 4 | 16

= 2516

98 x 18 = (9 x 1) + 8 | 8 x 8

= 9 + 8 | 64

= 1764

23 x 83 = (2 x 8) + 3 | 3 x 3

= 16 + 3 | 09

= 1909

58

= (5 x 5) + 8 | 8 x 8

= 25 + 8 | 64

= 3364

53

= (5 x 5) + 3 | 3 x 3

= 25 + 3 | 09

= 2809

How do you like these Vedic Maths technique, please let us know.

**A. Numbers whose last digits add to 10 and the remaining first digits are the same****Case 1**:**When sum of last two digits number is 100**Example:

**392 x 308**Here we can see that right digits sum is 100 i.e.(92 + 8) and left side digits are same. Here we can now apply the same method, which we discussed earlier for 2-digit number. But this time we must expect to have four figures on the right-hand side.

- First, multiply the right side numbers(92 x 08) and the result is 0736.
- Second, multiply 3 by the number that follows it, i.e.4, so the result of (3 x 4) is 12.
- And now the final output is
**120736**.

Example:

**795 x 705**Here 95 + 05 = 100 and left side digits are same i.e. '7'. Hence it qualifies for this case.

In calculation, we shall multiple the last two digits and the left digit i.e. '7; with its next number '8'. So the calculation is:

795 x 705 = 7 x 8 | 95 x 05

= 56 | 0475

=

**560475**Example:

**866 X 834**Here 66 + 34 = 100 and left side digit is 8 and its next number is 9. So the calculation is:

848 x 852 = 8 x 9 | 66 x 34 (Note: For 66 x 34, we shall discuss in our upcoming articles)

= 72 | 2244

=

**722244****Case 2: When sum of whose last digits is 10 and the remaining first digits are the same**Example:

**241 x 249**Here we can see that right digits sum is 10 i.e.(9 + 1) and left side digits are same i.e. 24. So we can now apply the same method as described above.

241 x 249 = 24 x 25 | 1 x 9

= 0600 | 09

=

**60009**444 x 446 = 44 x 45 | 4 x 6

= 1980 | 24

= 198025

147 x 143 = 14 x 15 | 7 x 3

= 0210 | 21

= 21021

**Special cases:****Case 1**: Where numbers are not qualified for above case in general, but can be qualified after converting their form. Example is given below.For example:

**93 x 39**93 × 39 doesn't seems to comes under this method. But if we convert 93 to (3 × 31), then 31 × 39 comes under this method. So the calculation is

93 x 39 = 3 x 31 x 39

= 3 x (31 x 39)

31 × 39 = 1209 (by same method discussed above i.e. 31 x 39 = (3 x 4) | (1 x 9) = 12 | 09 = 1209)

93 × 39 = 3 x 1209

=

**3627**The point to notice in the above example is that '39' needs '31' to qualify for method, and then we found that 93 is '3 × 31'.

**Case 2**: If the left hand digits are the same, but last digit's sum is not 10. We can still apply the same method that we discussed just above. For example:**35 x 37**In 35 x 37, both left hand numbers are 3, but 5 + 7 = 12 which is greater than 10. We can solve this in

**two ways.****First**, we modify the numbers in such a way that we can apply the method used for "squaring numbers ending in 5":35 x 37 = 35 x (35 + 2)

= (35 x 35) + (35 x 2)

= 1225 + 70 (Take 35 x 35 = 3 x 4 | 5 x 5 = 12 | 25 = 1225)

=

**1295****Second**approach is, we modify the numbers in such a way that we can apply the method discussed above. So we split 35 as 33 + 2. And observe that (33 x 37) comes under the above discussed method.So, we rewrite it as:

35 x 37 = (33 + 2) x 37

= (33 x 37) + (2 x 37)

= 1221 + 74 (Take 33 x 37 = (3 x 4) | (3 x 7) = 12 | 21 = 1221)

= 1295

One more example:

**114 x 117**= 114 x (116 + 1)= (114 x 116) + (114 x 1)

= 13224 + 114 (Take 114 x 116 = (11 x 12) | (4 x 6) = 132 | 24 = 13224)

=

**13338**OR

114 x 117 = (111 + 3) x 117 (Rest of the procedure is same)

Now, take one more example where first digit remains same but addition of last digits is less than 10.

68 x 61 = 68 x (62 - 1)

= (68 x 62) - (68 x 1) (Rest of the procedure is same)

OR

68 x 61 = (69 - 1) x 61

= (69 x 61) - (61 x 1) (Rest of the procedure is same)

**B.Numbers whose last digits are same and first digits add to 10**

**last figures are the same and the first figures add up to 10. This comes under the Vedic formula***"The First by the First and the Last by the Last"*. For example: 47 x 67Here we can see that left hand digits sum is 10 i.e.(4 + 6) and right hand digits are same i.e. 7 . So we can now apply the following method.

- Multiply the first digit of each number together i.e.(4 x 6 = 24). Add the last figure (7) into it (24 + 7 = 31) which is the first part of the answer.
- Multiplying the last figures together(i.e.7 × 7 = 49) which is the last part of the answer.
- The final answer is 3149.

Let us take some more examples

34 x 74 = (3 x 7) + 4 | 4 x 4

= 21 + 4 | 16

= 2516

98 x 18 = (9 x 1) + 8 | 8 x 8

= 9 + 8 | 64

= 1764

23 x 83 = (2 x 8) + 3 | 3 x 3

= 16 + 3 | 09

= 1909

**Note**: You can do "Squaring of numbers between 50 and 60" with the same method. You might be thinking that how this could be. But see, left hand digits sum is 10 i.e.(5 + 5) and right hand digits are same. So we can now apply the same method. We had already discussed it in "squaring numbers near 50", both are almost same but different presentation. See the following example:58

^{2}= 58 x 58= (5 x 5) + 8 | 8 x 8

= 25 + 8 | 64

= 3364

53

^{2}= 53 x 53= (5 x 5) + 3 | 3 x 3

= 25 + 3 | 09

= 2809

How do you like these Vedic Maths technique, please let us know.