Question

Find the least number which one division by 24, 36 and 48 leaves the remainder 21, 33 and 45 respectively.

Answer 1

Hint: Observe that 21 =24-3, 33 = 36 – 3 and 45 = 48 -3.Hence if a be the smallest number which on division by 21, 33 and 45 leaves remainder 0, then a-3 is the smallest number which on division by 24, 36 and 45 leaves remainder 21, 33 and 45.

Complete step-by-step answer:

Since 21 =24-3, 33 = 36 – 3 and 45 = 48 -3, if a is the smallest number which on division by 21, 33 and 45 leaves remainder 0, then a-3 is the smallest number which on division by 24, 36 and 45 leaves remainder 21, 33 and 45.
Let L be the LCM of 24,36 and 48. Hence from the definition of L, we have 24|L, 36|L and 48|L {a|b is read as a divides b}. Also let m be a natural number such that 24|M, 36|M and 48|M then we have $L\le M${Because M is a common multiple of 24,36 and 48 and L is least common multiple.}
Hence, we have a = L.
Now we know that
$\begin{align}
  & 24={{2}^{3}}\times 3 \\
 & 36={{2}^{2}}\times {{3}^{2}} \\
 & 48={{2}^{4}}\times 3 \\
\end{align}$
Hence, we have,
$L={{2}^{4}}\times {{3}^{2}}=144$
Hence a = 144.
Hence the smallest number which divides by 24,36 and 48 leaves remains 21, 33 and 45 respectively is a-3 = 144-3 = 141.

Note: [1] We will prove that if a is the smallest number which on division by 21, 33 and 45 leaves remainder 0, then a-3 is the smallest number which on division by 24, 36 and 45 leaves remainder 21, 33 and 45.
Proof:
As shown above a = L = LCM (24,36,45).
Since 24>3,36>3 and 45>3, we have L>3. So, a-3>0.
Let a-3 is not the smallest number with the above properties and let m be such a number.
So, we have mThen we have 24|m+3, 36|m+3 and 48|m+3.
Hence m+3 is a common multiple of 24,36 and 48 and hence $a\le m+3$
Hence, we have $a-3\le m$.
But, mHence, the assumption that a-3 is not the smallest number with the above properties is incorrect. Hence a- 3 is the smallest number with the above properties.