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uranmaximum [27]
3 years ago
8

Dado el número N = 4752a, averigua qué valor ha de tomar "a" para que: a) N sea divisible por tres. b) N sea divisible por cinco

. c) N sea divisible por 15. d) Cuando se divida N por 7 dé de resto 5.
Mathematics
1 answer:
lara [203]3 years ago
5 0

Answer:

I will answer it in English:

a) We have te number N = 4752a

if we want to divide it by 3, then the addition of all the digits must be a multiple of 3.

Sum = 4 + 7 + 5 + 2 + a = 6 + 7 + 5 + a = 18 + a

and 18 is a multiple of 3, then we can have, a = 0, a = 3, a = 6, a= 9

b) A number is divisible by 5 if it ends on 5 or 0, so the possible values of a are 0 and 5.

c) we know that 15*3 = 45

then 15*3000= 45000

now we have:

4752a - 45000 = 252a

15*13 = 2520

then we have that a must be equal to zero, and:

15*(13 + 3000) =  47520

d) We want a number that when we divide it by 7, the surpass is 5.

a number is divisible by 7 if the difference between the number without the units and the double of the units is 0 or a multiple of 7.

in N = 4752x the unit is a, then we have:

D = 4752 - 2*x

now, let's choose a such we have a multiple of 7

we know that 5746 is a multiple of 7, so we can choose x= 3 and get:

D = 4752 - 2*3 =   5746  

So the number N = 47523 is a multiple of 7, and when we divide it by 7 we will have a surpass of 0.

Then when we divide N + 5 by 7, we will have a surpass of 7.

This is: N + 5 = 47528

This means that the actual value of a that we need is 8.

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N is a positive integer
Murrr4er [49]

Part (1)

n is some positive integer. Let's say for now that n is even. So n = 2k, for some integer k

This means n-1 = 2k-1 is odd since subtracting 1 from an even number leads to an odd number.

Now multiply n with n-1 to get

n(n-1) = 2k(2k-1) = 2m

where m = k(2k-1) is an integer

The result 2m is even showing that n(n-1) is even

------------

Let's say that n is odd this time. That means n = 2k+1 for some integer k

And also n-1 = 2k+1-1 = 2k showing n-1 is even

Now multiply n and n-1

n(n-1) = (2k+1)(2k) = 2k(2k+1) = 2m

where m = k(2k+1) is an integer

We've shown that n(n-1) is even here as well.

------------

So overall, n(n-1) is even regardless if n is even or if n is odd.

Either n or n-1 will be even. If you multiply an even number with any number, the result will be even.

=======================================================

Part (2)

n is some positive integer

2n is always even since 2 is a factor of 2n

2n+1 is always odd because we're adding 1 to an even number. The sequence of integers goes even,odd,even,odd, etc and it does this forever.

-----------

Another way to see how 2n+1 is odd is to divide 2n+1 over 2 and you'll find that we get (2n+1)/2 = 2n/2+1/2 = n+0.5

The 0.5 at the end is not an integer, so there's no way that (2n+1)/2 is an integer; therefore 2n+1 is odd.

6 0
3 years ago
Evaluate each expression<br> 6! =<br> 3! • 2=<br> 31
Airida [17]
<h2><u>N-FACTORIAL!</u></h2>

<h3>\mathbb{  \color{brown} \: PROBLEM  }  \:   \:  \bold{{ \color{brown}{1}}} :</h3>

\mathcal{SOLUTION: }

  • \:   \bold{ \red{6!}= 6×5×4×3×2×1= \boxed{ \bold{ \blue{720}}}}

Therefore, <u>the value of 6! is 720</u>.

━┈─────────────────────┈━

<h3>\mathbb{  \color{brown} \: PROBLEM  }  \:   \:  \bold{{ \color{brown}{2}}} :</h3>

\mathcal{SOLUTION: }

  • \bold{3!  \cdot 2! }\implies  \bold{(3×2×1)  \cdot( 2×1 )}

  • \:   \: \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \implies \: \bold{ 6  \times 2} =  \boxed{ \bold{ \blue{12}}}

Therefore, <u>the value of the given expression is 12.</u>

━┈─────────────────────┈━

<h3><u>EXPLANATION</u><u>:</u></h3>
  • The mathematical symbol n! is read as "n factorial". The exclamation point "!" is read as factorial. And please remember or take note that 0! is equal to 1, and 1! is also equal to 1.

_______________∞_______________

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2n-3/5=5 solve for n
dezoksy [38]
N=14/5 thats the answer.
3 0
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Aleonysh [2.5K]

Answer:

Step-by-step explanation:

x+x+9=65

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x=28

28,37

4 0
3 years ago
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Answer:

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Step-by-step explanation:

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3 0
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