Find the smallest number which, when increased by 1 is exactly divisible by 12, 18, 24, 32, and 40.

1427

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1536

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1439

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2412

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 1439

Let us find out the L.C.M. of 12, 18, 24, 32, and 40

L.C.M.
= 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 1440

1440 = 1439 + 1

Hence, 1439 is the smallest number which, when increased by one is exactly divisible by the given numbers.

Suggest Corrections