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2 years ago

Find the greatest number that divides451, 328 and 673 leaving 1, 4 and 1,respectively as remainders.

Mathematics

1 answer:

Vesna [10]2 years ago

4 0

You're looking for the largest number x such that

x ≡ 1 (mod 451)

x ≡ 4 (mod 328)

x ≡ 1 (mod 673)

Recall that

x ≡ a (mod m)

x ≡ b (mod n)

is solvable only when a ≡ b (mod gcd(m, n)). But this is not the case here; with m = 451 and n = 328, we have gcd(m, n) = 41, and clearly

1 ≡ 4 (mod 41)

is not true.

So there is no such number.

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3 years ago

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2 years ago

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3 years ago

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A ball is thrown from a height of 38 meters with an initial downward velocity of 3 m/s

ANTONII [103]

<h2>Hello!</h2>

The answer is:

The ball will hit the ground after $t=2.47s$

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Since negative time does not exists, the ball will hit the ground after:

$t=2.473s=2.47s$

Have a nice day!

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