Question

Find the least number which is a multiple of 23, and on division by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.

Answer 1

Hint: Observe that 7 =18-11, 10 = 21 – 11 and 13 = 24 -11.Hence if a be the smallest number which on division by 18, 21 and 13 leaves remainder 0, then a-11 is the smallest number which on division by 18, 21 and 13 leaves remainder 7, 10 and 13. Hence if a is the smallest number satisfying the above property, then a-11 is a multiple of LCM(18,21,24), and a is a multiple of 23.

Complete step-by-step answer:
Since 7 =18-11, 10 = 21 – 11 and 13 = 24 -11. if a be the smallest number which on division by 18, 21 and 13 leaves remainder 0, then a-11 is the smallest number which on division by 18, 21 and 13 leaves remainder 7, 10 and 13.
Let L be the LCM of 18,21 and 24.
Now we know that
 $ \begin{align}
  & 18=2\times {{3}^{2}} \\
 & 21=3\times 7 \\
 & 24={{2}^{3}}\times 3 \\
\end{align} $
Hence, we have,
 $ L={{2}^{3}}\times {{3}^{2}}\times 7=504 $
Hence, we have L = 504
Since a+11 is a multiple of L, we have
 $ a=504k-11,k\in \mathbb{N} $
Also since a is a multiple of 23 and $ 504=21\times 23+21 $ , we have
 $ 21k-11 $ leaves remainder 0 on division by 23
We try to solve this using a hit and trial method.
When k = 1, we have
 $ 21k-11=10 $ , which leaves remainder 10 on division by 23
When k = 2, we have
 $ 21k-11=42-11=31=23\times 1+8 $ , which leaves remainder 8 on division by 23
When k = 3, we have
 $ 21k-11=63-11=52=23\times 2+6 $ , which leaves remainder 6 on division by 23.
When k = 4, we have
 $ 21k-11=84-11=73=23\times 3+4 $ , which leaves remainder 4 on division by 23.
When k = 5, we have
 $ 21k-11=105-11=94=23\times 4+2 $ , which leaves remainder 2 on division by 23.
When $ k=6 $ , we have
 $ 21k-11=126-11=115=23\times 5+0 $ , which leaves remainder 0 on division by 23.
Hence by hit and trial, we get k = 6.
Hence, we have the smallest number satisfying the above property is $ 504\times 6-11=3013 $
Hence the smallest number which is a multiple of 23 and on division by 18,21 and 24 leaves remainders 7, 10 and 13 respectively is 3013

Note: Verification:
We can verify the correctness of our solution by checking that 3013 leaves remainders 7,10 and 14 respectively on division by 18,21 and 24 and 3013 is also a multiple of 23
We have
 $ 3013=18\times 167+7 $
Hence, we have 3013 leaves remainder 7 on division by 18
 $ 3013=21\times 143+10 $
Hence, we have 3013 leaves remainder 10 on division by 21
 $ 3013=24\times 125+13 $
Hence, we have 3013 leaves remainder 13 on division by 24
 $ 3013=23\times 121 $
Hence, we have 3013 is a multiple of 23
Hence our solution is verified to be correct.