What is the largest number that divides evenly into the sum of any four consecutive numbers?

Answer 1

Well this is a bit of a trck question

All the above answers are correct but the LARGEST number will always be the sum of the numbers

n + (n + 1) + (n + 2) + (n + 3)

= 4n + 6

= 2(2n + 3)

= 2(n + (n + 3))

= 1(4n + 6)

ie the factors of 4n + 6 are 1, 2, 2n + 3 and 4n + 6

Note the common factors for all such sums are 1 and 2.

Answer 2

Sum = n + (n+1) + (n+2) + (n+3) + (n+4) Sum = 5n + (1+2+3+4) Sum = 5n + 10 Sum = 5 (n + 2) Thus the only number which always divides the sum of any five consecutive whole numbers is 5. I don't know what "the largest" means. If n≥3 then "the largest" = n+2 otherwise "the largest" = 5

Answer 3

Note that the sum of four consecutive numbers is the sum of two even numbers and two odd numbers, so must be even. Can a number bigger than two divide every sum of four consecutive numbers? Let's do some examples.

1+2+3+4=10, 2+3+4+5=14. Two is the greatest common divisor of 10 and 14, therefore no greater number can divide these two in particular. So the answer is 2.

Answer 4

the first number plus the last number
1+2+3+4=10 (1+4 = 5)
99+100+101+102= 402 (99+102 = 201)
but if you want just one single number everytime it would be 2
cause no matter what 4 numbers you add it will be even

Answer 5

a+(a+1)+(a+2)+(a+3) = 4a+6 = 2(2a+3).
So the largest number that *always* works is 2.

Answer 6

I think this is a bit vague, can you elaborate so that we could come up with an equation ;-) because you'd have an infinite set of numbers here.

Answer 7

x + (x + 1) + (x + 2) + (x + 3) = 4x + 6

4x + 6 = 2(2x + 3)

Your answer would be 2x + 3