Question

The greatest number, which when divided by 12, 15, 20 and 54 leaves a remainder of 8 in each case.

• A. 548
• B. 1088
• C. 1620
• D. Can’t determined

Answer 1

The correct option is D Can’t determined

We have to find the L.C.M of 12, 15, 20 and 54
The prime factor of $12=2\times 2\times 3=2^2\times 3$
The prime factor of $15=3\times 5$
The prime factor of $20=2\times 2\times 5=2^2\times 5$
The prime factor of $54=2\times 3\times 3\times 3=2\times 3^3$
L.C.M = Product of each prime factor with highest index used.
$=2^2\times 3^3\times 5=540$
The least number which give 8 as the remainder is 548 $(540\times 1+8=548).$
Similarly, second number which 8 as the remainder is 1088 $(540\times 2+8=1088).$
For third number, it is 1620 $(540\times 3+8=1620).$
So, we can’t able to find the largest number as we have infinite multiples of 540.
Hence, we can’t determine the greatest number, which when divided by 12, 15, 20 and 54 leaves a remainder of 8 in each case.