Question

Find four consecutive integers such that 8 times the sum of the first and the third is 40 greater than 10 times the fourth

Answer 1

All four consecutive integers will be 9, 10, 11, and 12 and can be found out by high common sense.

What is the pattern?

A pattern is a specific arrangement of numbers or any term which will follow a specific code or hint.

For example 2,4,6,8 in this, the difference between any two-term is only 2 and this will be its specific pattern.

Given that,

8 times the sum of the first and the third  =  8( 1st + 3rd)

40 greater than 10 times the fourth = 40 + 10(4th)

So,

8( 1st + 3rd) = 40 + 10(4th)

Now if we look right and side then it will come out as any number with the unit placed as 0.

So it means on the left side also the unit place must be 0.

By thinking of  1st + 3rd = 10 /20 because at 45 it will be 40 which is not equal to the right-hand side.

Now by hit and trial method, we need to guess it 20 = 9 + 11

Then

8(9 + 11) = 40 + 10(12)

One equation can never give us three unknowns hence it will be the correct way.

Hence so the four consecutive integers will be 9, 10, 11, and 12.

For more information about the pattern

brainly.com/question/14013716

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