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2 years ago

Show that any number of the form ABCABC (e.g. 356356) is divisible by 13.

Mathematics

1 answer:

Shtirlitz [24]2 years ago

8 0

We could write such a number as

$10^5a+10^4b+10^3c+10^2a+10b+c$

$=10^2(10^3+1)a+10(10^3+1)b+(10^3+1)c$

$=(10^3+1)(100a+10b+c)$

$10^3+1=1001=13\cdot77$ so we're done.

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To describe a specific arithmetic sequence, Elijah wrote the recursive formula:

Rufina [12.5K]

Answer:

Y(n) = 7n + 23

Step-by-step explanation:

Given:

f(0) = 30

f(n+1) = f(n) + 7

For n=0 : f(1) = f(0) + 7

For n=1 : f(2) = f(1) + 7

For n=2 : f(3) = f(2) + 7 and so on.

Hence the sequence is an arithmetic progression with common difference 7 and first term 30.

We have to find a general equation representing the terms of the sequence.

General term of an arithmetic progression is:

T(n) = a + (n-1)d

Here a = 30 and d = 7

Y(n) = 30 + 7(n-1) = 7n + 23

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3 years ago

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Answer:

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2 years ago

An English teacher needs to pick 9 books to put on his reading list for the next school year, and he needs to plan the order in

Nikitich [7]

Answer: $4.37\times10^3$

Step-by-step explanation:

Given : Number of choices for novels = 4

Number of choices for plays = 6

Number of choices for poetry books = 5

Number of choices for nonfiction books = 5

Total books =4+6+5+5=20

If he wants to include all 4 novels, then the number of books left to select = 9-4=5

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Then, the number of different reading schedules are possible :_

$^4C_4\times^{16}C_5\\\\=\dfrac{4!}{4!(4-4)!}\times\dfrac{16!}{5!(16-5)!}\\\\=(1)\times\dfrac{16\times15\times14\times13\times12\times11!}{120\times11!}\\\\=4368=4.368\times10^3\approx4.37\times10^3$

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3 years ago

8x + 1 ≤ 9 + 6x ≤ 5 + 10x

Andreas93 [3]

Answer:

x = -19

Step-by-step explanation:

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2 years ago