Question

What is the largest number which divides 1475, 3155 and 5255 leaves the same remainder in each case?
(a) 320
(b) 420
(c) 350
(d) 410

Answer 1

Hint: We have been given three dividends. So, we can use the remainder formula, Dividend = divisor \[\times \] quotient + remainder. We take the largest number as ‘x’ and remainder as ‘r’. We then subtract the three equations formed from one another. We then find the HCF of the dividend thus formed which will be the divisor that divides all the three numbers leaving the same remainder. We then conclude the problem by verifying this.

Complete step by step answer:
We have been given three numbers 1475, 3155 and 5255. It is said that they leave the same remainder. Let the greatest number that will divide 1475, 3155 and 5255 be ‘x’ and let the remainder be ‘r’. By Number theory we know that,
Dividend = divisor \[\times \] quotient + remainder
Here we take a reminder as ‘r’, and a divisor as ‘x’. We have been given the dividend as 1475, 3155 and 5255. Let ${{Q}_{1}}$, ${{Q}_{2}}$ and ${{Q}_{3}}$ be the quotient of their 3 numbers.
Thus we can form 3 equations as,
$1475=\left( {{Q}_{1}}\times x \right)+r$ ---(1).
$3155=\left( {{Q}_{2}}\times x \right)+r$ ---(2).
$5255=\left( {{Q}_{3}}\times x \right)+r$ ---(3).
Now let us subtract (1) from (2).
Subtracting (1) from (2) we get,
   $3155=\left( {{Q}_{2}}\times x \right)+r$.
$-\underline{\left( 1475=\left( {{Q}_{1}}\times x \right)+r \right)}$.
\[1680=\left( \left( {{Q}_{2}}-{{Q}_{1}} \right)\times x \right)+0\].
\[1680=\left( \left( {{Q}_{2}}-{{Q}_{1}} \right)\times x \right)\].
Now let us subtract (2) from (3).
Subtracting (2) from (3) we get,
   $5255=\left( {{Q}_{3}}\times x \right)+r$.
$-\underline{\left( 3155=\left( {{Q}_{2}}\times x \right)+r \right)}$.
\[2100=\left( \left( {{Q}_{3}}-{{Q}_{2}} \right)\times x \right)+0\].
\[2100=\left( \left( {{Q}_{3}}-{{Q}_{2}} \right)\times x \right)\].
Now let us subtract (1) from (3).
Subtracting (1) from (3) we get,
   $5255=\left( {{Q}_{3}}\times x \right)+r$.
$-\underline{\left( 1475=\left( {{Q}_{1}}\times x \right)+r \right)}$.
\[3780=\left( \left( {{Q}_{3}}-{{Q}_{1}} \right)\times x \right)+0\].
\[3780=\left( \left( {{Q}_{3}}-{{Q}_{1}} \right)\times x \right)\].
Hence by using the number theory, we got three new equations as,
\[1680=\left( \left( {{Q}_{2}}-{{Q}_{1}} \right)\times x \right)\] ---(4).
\[2100=\left( \left( {{Q}_{3}}-{{Q}_{2}} \right)\times x \right)\] ---(5).
\[3780=\left( \left( {{Q}_{3}}-{{Q}_{1}} \right)\times x \right)\] ---(6)
We can clearly see that now we have 3 distinct numbers 1680, 3780 and 2100 such that x completely divides all the 3 numbers. You can either find them separately or there are three numbers together using prime factorization i.e. it is the process and making the number into a multiple of prime numbers.
$\begin{align}
  & 2\left| \!{\underline {\,
  1680,2100,3780 \,}} \right. \\
 & 2\left| \!{\underline {\,
  840,1050,1890 \,}} \right. \\
 & 3\left| \!{\underline {\,
  420,525,945 \,}} \right. \\
 & 5\left| \!{\underline {\,
  140,175,315 \,}} \right. \\
 & 7\left| \!{\underline {\,
  28,35,63 \,}} \right. \\
 & \text{ }\left| \!{\underline {\,
  4,5,9 \,}} \right. \\
\end{align}$
Hence the $HCF\left( 1680,2100,3780 \right)=2\times 2\times 3\times 5\times 7$.
$HCF\left( 1680,2100,3780 \right)=420$.
So, we got the HCF of 3 numbers as 420.
Hence we can write equation (4), (5) and (6) as,
$1680=420\times 4$.
$2100=420\times 5$.
$3780=420\times 9$.
Thus 420 is the largest divisor, which leaves the same remainder with 1475, 3155 and 5255.
$1475=\left( 420\times 3 \right)+215$.
$3155=\left( 420\times 7 \right)+215$.
$5255=\left( 420\times 12 \right)+215$.
As you can see that all the three numbers leave the same remainder 215 when divided with 420.
Hence the required divisor is 420.

So, the correct answer is “Option B”.

Note: We can also solve this without using the division formula as shown below. We just subtract the numbers that are given in the problem. The numbers we get after subtraction are 1680, 2100 and 3780 i.e.
\[3155-1475=1680\].
\[5255-3155=2100\].
\[5255-1475=3780\].
The HCF of these 3 numbers is the required largest divisor, which leaves the same remainder. We have already computed it as 420.