Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        find  how of 81 and 237 and express it as linear combination of 81 and 237 i.e 237x =31x+237y for some  x and y`
2 years ago

```							Since 237 > 81, we apply the division lemma to 237 and 81 to obtain237 = 81 x 2 + 75  Step 1 Since remainder 75 ? 0, we apply the division lemma to 81 and 75 to obtain81 = 75 x 1 + 6  Step 2 Since remainder 6 ? 0, we apply the division lemma to 75 and 6 to obtain75 = 6 x 12 + 3  Step 3 Since remainder 3 ? 0, we apply the division lemma to 6 and 3 to obtain6 = 3 x 2 + 0  Step 4 In this step the remainder is zero. Thus, the divisor i.e. 3 in this step is the H.C.F. of the given numbers The H.C.F. of 237 and 81 is 3 From Step 3:3 = 75  6 x 12  Step 5 From Step 2:6 = 81  75 x 1 Thus, from Step 5, it is obtained3 = 75  (81  75 X 1) x 12? 3 = 75  (81x 12  75 x 12)? 3 = 75 x 13  81x 12  Step 6 From Step 1;75 = 237  81 x 2 Thus, from Step 6;3 = (237  81 x 2) x 13  81X 12? 3 = (237 x 13  81 x 26)  81x 12? 3 = 237 x 13  81 x 38? H.C.F. of 237 and 81 = 237 x 13 + 81 x (38) The above relationship is the representation of H.C.F. of 237 and 81 as linear combination of these two. RegardsArun (askIITians forum expert)
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on 10 grade maths

View all Questions »  ### Course Features

• 728 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions