Question

The least number when divided by 36,49 and 64 leaves remainder 25, 37 and 53 respectively is
1) 515
2) 535
3) 565
4) 585

Answer 1

Hint: We can observe that we have to find the least number such that when it is divided by 36,48 and 64, it leaves the remainder that is exactly 11 less than the divisor. Thus we can find the L.C.M. of the numbers 36,48 and 64, and subtract 11 from it.

Complete step-by-step answer:
Here we can observe a relation between the divisor and the remainder.
The pattern is that the remainder is always 11 less than the divisor.
\[\begin{array}{*{20}{l}}
  {36 - 11 = 25} \\
  {48 - 11 = 37} \\
  {64 - 11 = 53}
\end{array}\]
Thus we conclude that the required dividend should be 11 less than the exact multiple of divisor.
For the three numbers 36,48 and 64, we can find the least common multiple by finding the L.C.M. of these numbers.
Factorising the divisors into powers of prime factors we get,
$
  36 = {2^2}{3^2} \\
  48 = {2^4}3 \\
  64 = {2^6} \\
$
The L.C.M. of the divisors will be the multiplication of the unique prime factors from all the different factorisations raised to the highest occurring power.
Here we have two unique prime factors that are 2 and 3.
The highest power of 2 is 6, in the prime factorisation of 64.
Similarly, the highest power of 3 is 2, in the prime factorisation of 36.
THE L.C.M. is therefore ${2^6} \times {3^2} = 576.$
The required dividend is 11 less than the L.C.M. of the divisors, therefore the required number is\[576 - 11 = 565\].

Thus the option C is correct.

Note: L.C.M. of a sequence of numbers gives the smallest number which is divisible by each of the numbers. The L.C.M. of the numbers will be the multiplication of the unique prime factors from all the different factorisations raised to the highest occurring power.