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| Math Central | Quandaries & Queries |
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Question from Joshua, a student: We were given a question that involved consecutive integers. I can not seem to solve this question and I am the top student in my class.The question states: Find four consecutive integers such that twice the sum of the two greater integers exceeds three times the first by 91. How do you solve this question? Thank you very much for your support. |
We have two responses for you
Joshua,
Have you tried letting the integers be n, n+1,n+2 & n+3? Then your statement implies 2(n+2+n+3) = 3n+91?
Penny
Hi Joshua.
They are consecutive integers, so call the lowest one L. Then the
others are L+1, L+2 and L+3.
"Twice the sum of the two greater integers..."
2(L+2 + L+3)
"...exceeds three times the first by 91."
= 3(L) + 91
So
2(L+2 + L+3) = 3L + 91.
Can you solve for L now?
Cheers,
Stephen La Rocque.
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