Question

The sum of two numbers is 1560 and their HCF is 65. Find the total number of such pair
of numbers possible.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• Two number have HCF as 65.

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{first \: number \: be \: 65x} \\ &\sf{second \: number \: be \: 65y} \end{cases}\end{gathered}\end{gathered}$

where x and y are coprime numbers ( means HCF of x and y is 1).

According to statement,

$\rm :\longmapsto\:65x + 65y = 1560$

$\rm :\longmapsto\:65(x + y) = 1560$

$\rm :\longmapsto\:x + y = \cancel{\dfrac{1560}{65}} \: = 24$

$\rm :\implies\:x + y = 24$

Hᴇɴᴄᴇ,

➢ Possible pair of numbers so that x and y are coprime and their sum is 24 are shown in the below table :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 1 & \sf 23 \\ \\ \sf 5 & \sf 19 \\ \\ \sf 7 & \sf 17\\ \\ \sf 13 & \sf 11 \\ \\ \end{array}} \\ \end{gathered}$

Hᴇɴᴄᴇ,

➢ Possible pair of numbers are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 65 & \sf 1495 \\ \\ \sf 325 & \sf 1235 \\ \\ \sf 455 & \sf 1105\\ \\ \sf 715 & \sf 845 \\ \\ \end{array}} \\ \end{gathered}$

Hᴇɴᴄᴇ,

➢ Total number of possible pairs is 4.

Additional Information :-

If x and y are two natural numbers having hcf 'h' and lcm 'l', then following hold :-

$\:\rm \: \longmapsto\:x \times y = h \times l$

$\rm :\longmapsto\:h \: \leqslant x \: or \: y$

$\rm :\longmapsto\:l \geqslant x \: or \: y$

$\rm :\longmapsto\:h \leqslant x \: or \: y \: \leqslant l$

$\rm :\longmapsto\:h \: is \: a \: factor \: of \: l$

Answer 2

Answer:

x = 260/11 , y = 16900/11

For expaination see the above image.