Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
2.0k views
in Mathematics by (600 points)
edited by
Find the smallest no which when increased by 17 is exactly divisible by 520 and 468.

Please log in or register to answer this question.

1 Answer

0 votes
by (266k points)

The given numbers are 520 and 468. 

The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting 17 from the 

LCM of 520 and 468. 

Prime factorisation of 520 = 2 × 2 × 2 × 5 × 13 

Prime factorisation of 468 = 2 × 2 × 3 × 3 × 13 

LCM of  520 and 468 = 2 × 2 × 2 × 3 × 3 × 5 × 13 = 4680. 

Smallest number which when increased by 17 is exactly divisible by both 520 and 468 = 4680 – 17 = 4663.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...