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The greatest number that divides 215,167 and 135 so as to leave the same remainder in each case

What is the approach to solve such questions

#gate2022
closed with the note: got it

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## 1 Answer

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Take the difference between all pair of these three numbers

$$215 – 167 = 48\\215 – 135 = 80\\167 – 135 = 32$$

Now take the GCD of the differences $$GCD(48, 80, 32) = 16$$
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I know how to solve

I just asked, the intuition behind it

can you explain?
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$215 = xD_1 + R$

$167 = xD_1 + R$

$135 = xD_3 + R$

We have to maximize $x$

$$x = \frac{48}{D_1 – D_2} = \frac{80}{D_1 – D_3} = \frac{32}{D_2 – D_3}$$

$$x = \frac{16\times 3}{D_1 – D_2} = \frac{16\times5}{D_1 – D_3} = \frac{16\times2}{D_2 – D_3}$$

Now if the system $$D_1 – D_2 = 3\\D_1 – D_3 = 5\\D_2 – D_3 = 2$$ is solvable then $x = 16$ would be our answer.

Now rank of the matrix

$$Rank({\begin{bmatrix}1 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & -1\end{bmatrix}}) = 2$$

So it has infinite solutions, so $x = 16$ is our answer.