Question

Find a number which when divided by 39 gives a remainder 16 and when divided by 56 gives remainder 27. How many such numbers are there.

Answer 1

Hint: the number when divided by 39 and 56 leaves the remainders
16 and 27 respectively. we will take x as the number which when divided by 39 and 56 leaves us with remainders 16 and 27 respectively. from this x we will make two equations since we have been given two separate parameters. the equation being ….
\[\begin{align}
  & x=39a+16........(1) \\
 & x=56b+27..........(2) \\
 & \\
\end{align}\]
where ‘a’ and ‘b’ are positive integers which are also the multiple of 39 and 56 with which the unknown number is getting divided leaving the remainders 16 and 27.


Complete step by step solution:
From Eqn. (1) and Eqn. (2).
\[\begin{align}
  & \begin{array}{*{35}{l}}
   39a+16\text{ }=\text{ }56b\text{ }+27. \\
   or,\text{ }39a\text{ }=\text{ }56b+11\ldots ..\left( 3 \right) \\
\end{array} \\
 & 39a\text{ }=\text{ }39b+\left( 17b+11 \right).\text{ },\text{ }let\text{ }\left( 17b+11 \right)\text{ }=\text{ }39p \\
\end{align}\]
, [as \[39b+\left( 17b+11 \right)\] is divisible by 39 , and p is also a positive integer.]
Now,
\[\begin{align}
  & 17b+11\text{ }=\text{ }39p\ldots \ldots .\left( 4 \right) \\
 & 17b\text{ }=\text{ }34p\text{ }+\left( 5p-11 \right)\text{ or }17b\text{ }=\text{ (17}\times \text{2)}p\text{ }+\left( 5p-11 \right) \\
 & \\
\end{align}\]
[thus (5p-11) should also be a multiple of 17.]
Putting
 \[\begin{align}
  & 5p-11=17\text{ } \\
 & \text{ }p\text{ }=\text{ }\dfrac{28}{5}\text{ }\!\![\!\!\text{ not acceptable }\!\!]\!\!\text{ } \\
 & \text{thus putting } \\
 & 5p-11\text{ }=\text{ 17}\times \text{2 } \\
 & \text{ }p=\text{ }\dfrac{45}{5}\text{ }=\text{ }9. \\
\end{align}\]
putting p=9 in Eqn. (4)
\[\begin{array}{*{35}{l}}
   17b+11=39\times 9. \\
   \begin{align}
  & \text{or }17b\text{ }=\text{ }35111 \\
 & \text{ or }b\text{ }=\dfrac{340}{17}=20. \\
\end{align} \\
\end{array}\]
Hence, the number \[x=56b+27=\text{ }56\times 20+27\text{ }=\text{ }1120+27\text{ }=\text{ }1147\text{ }\] is Answer.
Set of such numbers is \[\left( 1147+39\times 56\times n \right)\]
or\[\left( 1147+2184n \right)\], where n= 0 , 1 , 2 , 3 , 4 , ……n terms.


Note: Note that whenever we are solving these types of problems,first we need to form two equations having the same value of \[x\] which when divided by these two different numbers will yield two different remainders.