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alekssr [168]
3 years ago
9

Prove that every integer greater than 7 can be written by using 3's and 5's only. that is, for every n > 7 there exist non-ne

gative integers tn, sn such that n = 3tn + 5sn.
Mathematics
1 answer:
Taya2010 [7]3 years ago
4 0
An integer may be a multiple of 3.
An integer may be 1 greater than a multiple of 3.
An integer may be 2 greater than a multiple of 3.

It is redundant to say an integer is 3 greater than a multiple of 3 (that's just a multiple of 3, we've got it covered).  Same for 4, 5, 6, 7...

Let's consider a number which is a multiple of 3.  Clearly, we can write 3+3+3+3+... until we reach the number.  It can be written as only 3's.

Let's consider a number which is 2 greater than a multiple of 3.  If we subtract 5 from that number, it'll be a multiple of 3.  That means we can write the number as 5+3+3+3+3+...  Of course, the number must be at least 8.

Let's consider a number which is 1 greater than a multiple of 3.  If we subtract 5 from that number, it'll be 2 greater than a multiple of 3.  If we subtract another 5, it'll be a multiple of 3.  That means we can write the number as 5+5+3+3+3+3+...  Of course, the number must be at least 13.

That's it.  We considered all the numbers.  We forgot 9, 10, 11, and 12, but these are easy peasy.

Beautiful question.
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If AC=18x-7,find the x. Round to the nearest hundredth if necessary <br>AB=7x+29 <br>BC=30​
Margaret [11]

Answer:

x=6

Step-by-step explanation:

AC = AB + BC

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18x-7=7x+29+30

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18x-7+7=7x+59+7

18x=7x+66

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Therefore answer is 6200/604800= 0.01(approximated)

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Which of these statements is true?
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B all square are regular polygons

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