Divisibility and Multiples | FIO

Question 12

How many multiples of 9 are there between the numbers 4300 and 4400?

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Solution

We will find the first multiple of 9 after 4300 and the last multiple of 9 before 4400. Then we will count how many multiples are in this range.

Step 1 — Find the first multiple of 9

We want to find the smallest multiple of 9 that is greater than 4300. Let us divide 4300 by 9.

4300÷94300 \div 9

=477 with a remainder of 7= 477 \text{ with a remainder of } 7

This means 4300=9×477+74300 = 9 \times 477 + 7. The number 4300 is 7 more than 9×477=42939 \times 477 = 4293. So, the next multiple of 9 will be 9×(477+1)9 \times (477 + 1).

9×4789 \times 478

=4302= 4302

First multiple=4302\boxed{\text{First multiple} = 4302}

Step 2 — Find the last multiple of 9

We want to find the largest multiple of 9 that is less than 4400. Let us divide 4400 by 9.

4400÷94400 \div 9

=488 with a remainder of 8= 488 \text{ with a remainder of } 8

This means 4400=9×488+84400 = 9 \times 488 + 8. The number 4400 is 8 more than 9×488=43929 \times 488 = 4392. So, the largest multiple of 9 less than 4400 is 9×4889 \times 488.

9×4889 \times 488

=4392= 4392

Last multiple=4392\boxed{\text{Last multiple} = 4392}

Step 3 — Count the multiples

We have the first multiple as 4302 and the last multiple as 4392. These multiples form an arithmetic progression (a sequence where the difference between consecutive terms is constant). The common difference is 9. Let nn be the number of multiples. We can use the formula: Last multiple=First multiple+(n1)×common difference\text{Last multiple} = \text{First multiple} + (n-1) \times \text{common difference}.

4392=4302+(n1)×94392 = 4302 + (n-1) \times 9

43924302=(n1)×94392 - 4302 = (n-1) \times 9

90=(n1)×990 = (n-1) \times 9

909=n1\frac{90}{9} = n-1

10=n110 = n-1

n=10+1n = 10 + 1

n=11n = 11

Number of multiples=11\boxed{\text{Number of multiples} = 11}

Answer

(i) There are 11 multiples of 9 between 4300 and 4400.

More questions in FIO

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