The least multiple of 13, which on dividing by 4,5, 6,7 and 8 leaves remainder 2 in each case is

India's Super Teachers for all govt. exams Under One Roof

Enroll For Free Now

Given:

Multiple of 13 divided by 3, 4, 5, and 6 gives remainder 1, 2, 3, and 4.

Concept used:

LCM of number.

Calculation:

The difference between the divisor and the corresponding remainder is the same in each case

i.e 3 – 1 = 2, 4 – 2 = 2, 5 – 3 = 2, 6 – 4 = 2

⇒ LCM(3, 4, 5, 6)

⇒ LCM = 2 × 3 × 2 × 5

⇒ LCM = 60

Let the required number be 60k – 2 which is a multiple of 13.

The least value of ‘k’ for which (60k – 2) is divisible by 13 is k = 10.

⇒ required number = 60 × 10 – 2

⇒ 598

∴ 598 is the least multiple of 13 divisible by 3, 4, 5, and 6 gives remainder 1, 2, 3, and 4.

India’s #1 Learning Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Get Started for Free Download App

Trusted by 3.4 Crore+ Students

37.

The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35, is

B.

4714

L.C.M. of 20, 28, 32, 35 = 1120
Hence, the required number = 5834 -  1120 = 4714

Home » Aptitude » LCM and HCF » Question

1. The least multiple of 13, which on dividing by 4, 5, 6, 7 and 8 leaves remainder 2 in each case is:

According to question , LCM of 4, 5, 6, 7 and 8

LCM of 4, 5, 6, 7 and 8 = 2 × 2 × 2 × 3 × 5 × 7 = 840.Let required number be 840K + 2 which is multiple of 13.Least value of K for which ( 840K + 2 ) is divisible by 13 is K = 3∴ Required number = 840 × 3 + 2

Required number = 2520 + 2 = 2522