abhi7605 wrote this blog titled "VEDIC MATHS - MULTIPLICATION BY SERIES OF 9 caseIII"

MULTIPLICATION BY SERIES OF 9

CASE III

WHEN 9 SERIES IS LESS THAN NO.

STEP 1.COUNT THE NUMBER OF 9 IN THE SERIES AND MARK THAT IN THE NO.

STEP 2. Reduce the no. by 1 + the remaining no without mark

STEP 3. Take complementary of the marked no.

STEP 4. Write the final answer

e.g. 1 - 341 x 99

STEP 1.COUNT THE NUMBER OF 9 IN THE SERIES AND MARK THAT IN THE NO.

341 x 99 (there r 2 9’s so we have taken 2 nos. )

STEP 2. Reduce the no. by 1+ the remaining no without mark

3 + 1 =4

341-4=337

STEP 3. Take complementary of the marked no.

Complementary of 41 is 59

STEP 4. 33759

Finally answer will be 33759

e.g 2 - 4178 x 999

STEP 1.COUNT THE NUMBER OF 9 IN THE SERIES AND MARK THAT IN THE NO.

4178 x 999

(there r 3 9’s so we have taken 3 nos. )

STEP 2. Reduce the no. by 1+ the remaining no without mark

4 + 1 =5

4178-5=4173

STEP 3. Take complementary of the marked no.

Complementary of 178 is 822

STEP 4. 4173822

Finally answer will be 4173822

e.g 3 - 50132 x 9999

STEP 1.COUNT THE NUMBER OF 9 IN THE SERIES AND MARK THAT IN THE NO.

50132 x 9999

(there r 4 9’s so we have taken 4 nos. )

STEP 2. Reduce the no. by 1+ the remaining no without mark

5 + 1 =6

50132-6=50126

STEP 3. Take complementary of the marked no.

Complementary of 0132 is 9868

STEP 4. 501269868

Finally answer will be 501269868