NMTC PRACTICE
Number Theory
Level
3
A number N leaves the same remainder while dividing 5814, 5430, 5958 . what is the largest possible value of N ???
The answer is 48.
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Let p, q, r and s be any number from the question, if r in remainder.
5814 = p X + r …………………. (i)
5430 = q X + r …………………. (ii)
5958 = s X + r ………………….. (iii)
from (i) & (ii)
384 = (p-q) X
from (ii) & (iii)
5430 – 5958 = (q - s) X
528 = (s - q) X
from (iii) & (i)
5814 – 5958 = (p - s) X
144 = (s - p) X
so we get three equation
384 = (p - q) X
528 = (s - q) X
144 = (s - p) X
(p - q) X = 2x2x2x2x2x2x2x3
(s - q) X = 2x2x2x2x x3x 11
(s - q) X = 2x2x2x2x x3x x3
So the HCF of these three numbers
= 2x2x2x2x3
= 48
So the required largest number is 48
Check:
48 x 121 = 5808 then + 6 = 5814
48 x 113 = 5424 then + 6 = 5430
48 x 124 = 5952 than + 6 = 5958