What number divided by 2 3 4 5 6 has a remainder of 1 but when divided by 7 has no remainder?

1 answer:

3 0

301

We could start by finding the lowest common multiple of 2, 3, 4, 5, and 6, which is 60. Then, we can consider the next few multiples: 120, 180, 240, 300...

However, because we need a remainder of 1 when our number is divided by each of these numbers (2,3,4,5,6), we want to go one above each of these multiples. So we're talking about 61, 121, 181, 241, 301... Those are the numbers that will satisfy the "remainder of 1" part of the question.

Now, we need to find out which one satisfies the other part of the question, which just requires dividing each of these numbers by 7 to see which is divisible by 7 (in other words, which one gives us a remainder of zero when we divide by 7).

301 does it. 301/7 = 43. So 301 is a multiple of 7 and therefore will yield no remainder when divided by 7.

Hope this all makes sense.

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Firlakuza [10]

Answer:

Step-by-step explanation:

6 0

2 years ago

3^x= 3*2^x solve this equation

kompoz [17]

In the equation

$3^x = 3\cdot 2^x$

divide both sides by $2^x$ to get

$\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3$

Take the base-3/2 logarithm of both sides:

$\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}$

Alternatively, you can divide both sides by $3^x$ :

$\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13$

Then take the base-2/3 logarith of both sides to get

$\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}$