Question

When four consecutive integers are added sum of 46 find tha integers

Answer 1

Let the common multiple be x

So, let the four consecutive integers be x, x + 1, x + 2 and x + 3


According to the given condition

x + x + 1 + x + 2 + x + 3 = 46

☛ 4x + 6 = 46

☛ 4x = 46 - 6

☛ 4x = 40

Dividing both the sides by 4

∴ x = 10


∴ 1st integer ☛ x

☛ $\pink{\underline{\pink{\underline{\tt{\red{10}}}}}}$


∴ 2nd integer ☛ x + 1

☛ 10 + 1

☛ $\pink{\underline{\pink{\underline{\tt{\red{11}}}}}}$


∴ 3rd integer ☛ x + 2

☛ 10 + 2

☛ $\pink{\underline{\pink{\underline{\tt{\red{12}}}}}}$


∴ 4th integer ☛ x + 3

☛ 10 + 3

☛ $\pink{\underline{\pink{\underline{\tt{\red{13}}}}}}$

Answer 2

Question :

When four consecutive integers are added sum of 46 find tha integers?

Answer :

Let the first consecutive number = x

2 nd number = x+1

3 rd number = x+2

4 th number = x+3

According to question ,

x+x+1+x+2+x+3 = 46

4x+6 = 46

4x = 40

x = 10

The required numbers are :

x = 10

x+1 = 11

x+2 = 12

x+3 = 13