Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Book Online Demo
Check Your IQ
Try Test

Courses

Dropper NEET CourseDropper JEE CourseClass - 12 NEET CourseClass - 12 JEE CourseClass - 11 NEET CourseClass - 11 JEE CourseClass - 10 Foundation NEET CourseClass - 10 Foundation JEE CourseClass - 10 CBSE CourseClass - 9 Foundation NEET CourseClass - 9 Foundation JEE CourseClass -9 CBSE CourseClass - 8 CBSE CourseClass - 7 CBSE CourseClass - 6 CBSE Course
Offline Centres

Q.

Find the HCF of  506 and 1155 and express it as a linear combination of them.


see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

11;11=  506 × 16 - 1155 × 7

b

12;12=  506 × 16 - 1155 × 6

c

12;12=  506 × 16 - 1155 × 7

d

13;13=  506 × 16 - 1155 × 7 

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given to find the HCF of two integers 506 and 1155 and to express it as linear combination of 506 and 1155.
Euclid's Division Lemma states that if two positive integers a and b exist, then there must be unique values of q and r that satisfy the formula a= bq + r, where  0 ≤ r < b.
Apply Euclid’s division lemma on 506, 1155   . . .  . . .(1)
1155=506×2+143  
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 506 and remainder 143
506=143×3+77   . . . . . . (2)
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 143 and remainder 77
143=77×1+66   . . . . . . (3)
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 77 and remainder 66
77=66×1+11   . . . . . . .(4)
∵  remainder ≠ 0,  apply Euclid’s division lemma on divisor 66 and remainder 11
66=11×6+0   . . . . . .(5)
∴   HCF of 506, 1155 is 11
Now,from (4),
11=77−66×1   . . . . . .(6)
Substitute (3) in (6)
11=77−(143−77×1)×1 ⇒77−143×1+77×1 ⇒77×2−143×1             . . . . . . . . .(7)  
Substitute (2) in (7),
11=(506−143×3)×2−143×1 ⇒506×2−143×6−143×1 ⇒506×2−143×7                  … . . . . ..(8)  
Substitute (1) in (8),
11=506×2−(1155−506×2)×7 ⇒506×16−1155×7 ⇒506×2−1155×7+506×14  
∴  HCF of 506 and 1155 is 11 and can be expressed as a linear combination of 506 and 1155 by,
11=  506 × 16 - 1155 × 7.
Hence the correct option is 1.
 
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon
Find the HCF of  506 and 1155 and express it as a linear combination of them.