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

The smallest positive integer n' with 24 divisors (where 1 and n are also considered as divisors of n) is

A
420
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
240
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
360
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
480
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 360
Letusinvestigatetheoptionsaspertheguidelinesofthegivenpassage.OptionA420=22×31×231a=2,b=1,c=1numberofdivisors=(2+1)(1+1)(1+1)=1224.OptionB240=24×31×51a=4,b=1,c=1numberofdivisors=(4+1)(1+1)(1+1)=2024.OptionC360=23×32×51a=3,b=2,c=1numberofdivisors=(3+1)(2+1)(1+1)=24.OptionD480=25×31×51a=5,b=1,c=1numberofdivisors=(5+1)(1+1)(1+1)=24.BothoptionCandDcomplieswiththegivenguidence.ButoptionCi.e360islesser.AnswerOptionC

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Lowest Common Multiple
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon