Question

find the greatest number which on being divided by 9, 12, 24, 45 leaves 3, 6, 18 and 39 as remainder​

Answer 1

Answer:

Consider the following numbers which are divisible by 9, using the test of divisibility by 9: 99, 198, 171, 9990, 3411. Sum of the digits of 99 = 9 + 9 = 18, which is divisible by 9. Hence, 99 is divisible by 9.

Answer 2

Answer:

First of all let us find the LCM of 9,12,24 and 45

LCM (9,12,24,45)=2^2*3^2*5=360

now 9–3=6,

2–6=6,

24–18=6,

45–39=6

all time the remainder is divisor-6

Thus the number is in the form of

360*n -6

now the biggest 5 digit number is 99999

and 99999=277*360+279

thus our require number is

=277*360–6 = 99714