Asked by Mo
the sum of 5 consecutive whole numbers is 2005. what is the sum of all the digits of these 5 numbers
Answers
Answered by
saiko
so you have,
(n-2)+(n-1)+n+(n+1)+(n+2)=2005
5n=2005
n=401
nos are,
399,400,401,402,403
(n-2)+(n-1)+n+(n+1)+(n+2)=2005
5n=2005
n=401
nos are,
399,400,401,402,403
Answered by
Damon
n-4 + n-2 + n + n+2 + n+4 = 2005
5 n = 2005
n = 401
so the list
397
399
401
403
405
so
2*3 + 3*4 + 2*9 + 25
= 61
5 n = 2005
n = 401
so the list
397
399
401
403
405
so
2*3 + 3*4 + 2*9 + 25
= 61
Answered by
Damon
I misread it, thought it said ODD numbers
Answered by
Abigail
I’m trying to study and I have to 4 choices none of these are the answer
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