# Thread: " Help me in solving problem"

1. ## " Help me in solving problem"

I did not understand the given problem statement.
Could you please explain it clearly?
And give me step by step answer

The given problem is

" Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively."

Thanking you , advance.

Please explain in details.

2. Originally Posted by SEK790
I did not understand the given problem statement.
Could you please explain it clearly?
And give me step by step answer

The given problem is

" Find the greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively."

Thanking you , advance.

Please explain in details.
Are you sure this problem should be in Pre-Algebra and Algebra? This is a problem for the Chinese Remainder Theorem.

$\displaystyle x\equiv 6\ (\text{mod}\ 1657)$

$\displaystyle x\equiv 5\ (\text{mod}\ 2037)$

1657 and 2035 are pairwise coprime. Solving the CRT gives

$\displaystyle x \equiv 1305722\ (\text{mod}\ 3375309)$

So the answer is 1305722.

EDIT: I misread "greatest number" as "smallest positive integer." There are infinitely many numbers satisfying the given constraints, and there is no greatest number.

EDIT 2: Ah, it seems I made another misinterpretation. I thought you meant "dividing by" instead of "dividing." Okay so we have

$\displaystyle 1657\equiv 6\ (\text{mod}\ x)$

$\displaystyle 2037\equiv 5\ (\text{mod}\ x)$

$\displaystyle 1651\equiv 0\ (\text{mod}\ x)$

$\displaystyle 2032\equiv 0\ (\text{mod}\ x)$

where

$\displaystyle 1651 = 13\cdot 127$

$\displaystyle 2032 = 2^4\cdot 127$

So the answer is 127.

3. Since you haven't responded, and since the problem is easier than the problem I thought it was at first, I'll write a simpler explanation here.

We are after a number, call it x.

1657 divided by x gives remainder 6. That means that 1657 - 6 = 1651 is a multiple of x.

2037 divided by x gives remainder 5. That means that 2037 - 5 = 2032 is a multiple of x.

We take the greatest common divisor of 1651 and 2032, and to help with that I wrote the prime factorisations in my post above.

If the greatest common divisor had been less than or equal to 6, this problem would have no solution.