Math / Numbers & Operations

Question```latex Rastavite svaki od sledećih brojeva na proste činilace, a zatim pronađite dva uzastopna prirodna broja čiji proizvod iznosi:
\begin{enumerate} \item 12 \item 20 \item 42 \item 1040400 \item 72 \item 156 \item 600 \end{enumerate} ```

Studdy Solution

STEP 1

What is this asking? Find two whole numbers right next to each other that multiply to give us each of these bigger numbers. Watch out! It's easy to mix up factors and multiples, so make sure you're finding numbers that multiply to the given number, not numbers that the given number divides into.

STEP 2

1. Prime Factorization
2. Find the Consecutive Numbers

STEP 3

Alright, future mathematicians!
Let's start by breaking down each of these numbers into their prime factors.
This is like finding the secret code of each number!

STEP 4

$12 = 2 \cdot 2 \cdot 3 = 2^2 \cdot 3$ .
See how we found the smallest prime numbers that multiply together to make 12?

STEP 5

$20 = 2 \cdot 2 \cdot 5 = 2^2 \cdot 5$ . Two and five are the prime ingredients for 20!

STEP 6

$42 = 2 \cdot 3 \cdot 7$ .
A mix of two, three, and seven gives us 42!

STEP 7

$72 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 = 2^3 \cdot 3^2$ . 72 is made up of a bunch of twos and threes!

STEP 8

$1040400 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 11 = 2^4 \cdot 3^2 \cdot 5^2 \cdot 7 \cdot 11$ .
Wow, that's a big one, but we conquered it!

STEP 9

$156 = 2 \cdot 2 \cdot 3 \cdot 13 = 2^2 \cdot 3 \cdot 13$ . 156 has a little bit of everything!

STEP 10

$600 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 = 2^3 \cdot 3 \cdot 5^2$ . 600 is a party of twos, threes, and fives!

STEP 11

Now, let's find those consecutive numbers.
We're looking for two numbers right next to each other that, when multiplied, give us our original number.

STEP 12

$3 \cdot 4 = 12$ . Three and four are our winning pair!

STEP 13

$4 \cdot 5 = 20$ . Four and five are a perfect match!

STEP 14

$6 \cdot 7 = 42$ . Six and seven team up to make 42!

STEP 15

$8 \cdot 9 = 72$ . Eight and nine are consecutive and multiply to 72!

STEP 16

$1020 \cdot 1021 = 1040400$ .
Even for this large number, we found the consecutive pair: 1020 and 1021!

STEP 17

$12 \cdot 13 = 156$ . Twelve and thirteen work together to make 156!

STEP 18

$24 \cdot 25 = 600$ . 24 and 25 are our final consecutive duo!

STEP 19

We found the consecutive numbers for each given product!
For 12 it's 3 and 4, for 20 it's 4 and 5, for 42 it's 6 and 7, for 72 it's 8 and 9, for 1040400 it's 1020 and 1021, for 156 it's 12 and 13, and for 600 it's 24 and 25!

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Question```latex Rastavite svaki od sledećih brojeva na proste činilace, a zatim pronađite dva uzastopna prirodna broja čiji proizvod iznosi:\begin{enumerate} \item 12 \item 20 \item 42 \item 1040400 \item 72 \item 156 \item 600 \end{enumerate} ```

Studdy Solution

STEP 1

STEP 2

1. Prime Factorization2. Find the Consecutive Numbers

STEP 3

Alright, future mathematicians!Let's **start** by breaking down each of these numbers into their prime factors.This is like finding the secret code of each number!

STEP 4

12=2⋅2⋅3=22⋅312 = 2 \cdot 2 \cdot 3 = 2^2 \cdot 312=2⋅2⋅3=22⋅3.See how we found the smallest prime numbers that multiply together to make **12**?

STEP 5

20=2⋅2⋅5=22⋅520 = 2 \cdot 2 \cdot 5 = 2^2 \cdot 520=2⋅2⋅5=22⋅5. **Two** and **five** are the prime ingredients for **20**!

STEP 6

42=2⋅3⋅742 = 2 \cdot 3 \cdot 742=2⋅3⋅7.A mix of **two**, **three**, and **seven** gives us **42**!

STEP 7

72=2⋅2⋅2⋅3⋅3=23⋅3272 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 = 2^3 \cdot 3^272=2⋅2⋅2⋅3⋅3=23⋅32. **72** is made up of a bunch of **twos** and **threes**!

STEP 8

STEP 9

156=2⋅2⋅3⋅13=22⋅3⋅13156 = 2 \cdot 2 \cdot 3 \cdot 13 = 2^2 \cdot 3 \cdot 13156=2⋅2⋅3⋅13=22⋅3⋅13. **156** has a little bit of everything!

STEP 10

600=2⋅2⋅2⋅3⋅5⋅5=23⋅3⋅52600 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 = 2^3 \cdot 3 \cdot 5^2600=2⋅2⋅2⋅3⋅5⋅5=23⋅3⋅52. **600** is a party of **twos**, **threes**, and **fives**!

STEP 11

Now, let's find those consecutive numbers.We're looking for two numbers right next to each other that, when multiplied, give us our original number.

STEP 12

3⋅4=123 \cdot 4 = 123⋅4=12. **Three** and **four** are our winning pair!

STEP 13

4⋅5=204 \cdot 5 = 204⋅5=20. **Four** and **five** are a perfect match!

STEP 14

6⋅7=426 \cdot 7 = 426⋅7=42. **Six** and **seven** team up to make **42**!

STEP 15

8⋅9=728 \cdot 9 = 728⋅9=72. **Eight** and **nine** are consecutive and multiply to **72**!

STEP 16

1020⋅1021=10404001020 \cdot 1021 = 10404001020⋅1021=1040400.Even for this large number, we found the consecutive pair: **1020** and **1021**!

STEP 17

12⋅13=15612 \cdot 13 = 15612⋅13=156. **Twelve** and **thirteen** work together to make **156**!

STEP 18

24⋅25=60024 \cdot 25 = 60024⋅25=600. **24** and **25** are our final consecutive duo!

STEP 19

We found the consecutive numbers for each given product!For 12 it's 3 and 4, for 20 it's 4 and 5, for 42 it's 6 and 7, for 72 it's 8 and 9, for 1040400 it's 1020 and 1021, for 156 it's 12 and 13, and for 600 it's 24 and 25!

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Question```latex Rastavite svaki od sledećih brojeva na proste činilace, a zatim pronađite dva uzastopna prirodna broja čiji proizvod iznosi:
\begin{enumerate} \item 12 \item 20 \item 42 \item 1040400 \item 72 \item 156 \item 600 \end{enumerate} ```

1. Prime Factorization
2. Find the Consecutive Numbers

Alright, future mathematicians!
Let's start by breaking down each of these numbers into their prime factors.
This is like finding the secret code of each number!

$12 = 2 \cdot 2 \cdot 3 = 2^2 \cdot 3$ .
See how we found the smallest prime numbers that multiply together to make 12?

$20 = 2 \cdot 2 \cdot 5 = 2^2 \cdot 5$ . Two and five are the prime ingredients for 20!

$42 = 2 \cdot 3 \cdot 7$ .
A mix of two, three, and seven gives us 42!

$72 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 = 2^3 \cdot 3^2$ . 72 is made up of a bunch of twos and threes!

$156 = 2 \cdot 2 \cdot 3 \cdot 13 = 2^2 \cdot 3 \cdot 13$ . 156 has a little bit of everything!

$600 = 2 \cdot 2 \cdot 2 \cdot 3 \cdot 5 \cdot 5 = 2^3 \cdot 3 \cdot 5^2$ . 600 is a party of twos, threes, and fives!

Now, let's find those consecutive numbers.
We're looking for two numbers right next to each other that, when multiplied, give us our original number.

$3 \cdot 4 = 12$ . Three and four are our winning pair!

$4 \cdot 5 = 20$ . Four and five are a perfect match!

$6 \cdot 7 = 42$ . Six and seven team up to make 42!

$8 \cdot 9 = 72$ . Eight and nine are consecutive and multiply to 72!

$1020 \cdot 1021 = 1040400$ .
Even for this large number, we found the consecutive pair: 1020 and 1021!

$12 \cdot 13 = 156$ . Twelve and thirteen work together to make 156!

$24 \cdot 25 = 600$ . 24 and 25 are our final consecutive duo!

We found the consecutive numbers for each given product!
For 12 it's 3 and 4, for 20 it's 4 and 5, for 42 it's 6 and 7, for 72 it's 8 and 9, for 1040400 it's 1020 and 1021, for 156 it's 12 and 13, and for 600 it's 24 and 25!