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[logicxonomy]

Successive Remainder problems (LCM and HCF Short tricks)

Que: If N is divided by 3,4,5 and 6 successively it leaves remainders 2,1,1 and 4 respectively. How many such 4-digit numbers are possible?

(a) 100

(b) 144

(c) 150

(d) 176

Solution:

The sequence of divisors: 3, 4, 5, and then 6

The sequence of respective remainders: 2, 1, 1, and 4

Short Trick: From right to left apply this formula

(Divisor+Remainder)×Previous Divisor= New Divisor

(6+4)×5=50

(50+1)×4=204

(204+1)×3=615

(615+2)=617 …….The End

The least Possible number that satisfies this condition is 617.

The General formula of such numbers= LCM(3,4,5,6)×k +617

=60×k +617

Here k is a Whole number.

For least four-digit number

60k +617> 1000

k>6.38

For the largest four-digit Number

60k +617< 9999

k<156.36

It means k={7,8,9……,156}⇒150 numbers

So, a total of 150 such numbers exist.

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