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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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