Question

the product of three consecutive numbers is 74046. what is the sum of these numbers​

Answer 1

Numbers

Given:

3 consecutive numbers, whose product is 74046

To find:

Sum of these numbers.

Explanation:

Let the numbers be $x-1, x, x+1$

According to question,

$(x-1)\times (x)\times(x+1)=74046$

Here we need to find the consecutive factors of 74046

Because solving it using equation, will be quite tough, So using factorization method is easier in these type of questions.

Factors of 74046 are 2, 3 ,7, 41, 43

$74046=2\times3\times7\times41\times43$

$74046=41\times42\times43$

So, these three numbers are consecutive and also the factors of given number.

hence, x is 42.

So, sum of these numbers are

$(x-1+x+x+1)=3x\\\\=3\times42\\\\=126$

Hence, the sum is 126.

Answer 2

The product of three consecutive numbers is 74046 then the sum of these numbers​ is 126

Given:

• The product of three consecutive numbers is 74046

To Find:

• Sum of these numbers

Solution:

• "Prime Factorization is finding prime numbers/factors which when multiplied together results in the original number"
• Prime number is a natural number which has only two factors one and number itself.  (e.g. , 2 , 3 , 5 , 7 .... )

Step 1:

Prime factorize 74046

74046 = 2 ×  3 × 7 × 41 × 43

Step 2:

Multiply 2 , 3 and 7

74046 = 42 × 41 × 43

Step 3:

Rearrange the order

74046 =  41 × 42 × 43

41 , 42 and 43 are  three consecutive numbers

Step 4:

Calculate the sum

41 + 42 + 43 = 126

The product of three consecutive numbers is 74046 then the sum of these numbers​ is 126