CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

What is the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, and 7 on division by 8?

A
1679
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1039
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
839
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
979
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 839
Let the number be N.
When a number is divided by 5, a remainder of 4 can be thought of as a remainder of 1.
So, N=5a1 or N+1=5a
Similarly,
N=6b1 or N+1=6b
N=7c1 or N+1=7c
N=8d1 or N+1=8d
N+1 can be expressed as a multiple of (5,6,7,8)
Now, smallest value of N+1= LCM of (5,6,7,8)
Smallest value of N= LCM of (5,6,7,8)1
=8401=839

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Formation of Algebraic Expressions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon