SOLUTION: find the triple of consecutive integers such that 11 times the largest of the integers is 46 more than the product of the other two

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Question 567957: find the triple of consecutive integers such that 11 times the largest of the integers is 46 more than the product of the other two
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the integers +n+, +n%2B1+, and +n%2B2+
given:
+11%2A%28+n%2B2+%29+=+n%2A%28+n%2B1+%29+%2B+46+
+11n+%2B+22+=+n%5E2+%2B+n+%2B+46+
+n%5E2+-+10n+%2B+24+=+0+
complete the square:
+n%5E2+-+10n+%2B+%28-10%2F2%29%5E2+=+-24+%2B+%28-10%2F2%29%5E2+
+%28+n+-+5+%29%5E2+=+-24+%2B+25+
+n+-+5+=+1+
+n+=+6+
also,
+n+-+5+=+-1+
+n+=+4+
------------
If the numbers are:
+n+=+6+
+n+%2B1+=+7+
+n+%2B+2+=+8+
+11%2A%28+n%2B2+%29+=+n%2A%28+n%2B1+%29+%2B+46+
+11%2A8+=+6%2A7+%2B+46+
+88+=+42+%2B+46+
+88+=+88+
OK
If the numbers are:
+n+=+4+
+n+%2B+1+=+5+
+n+%2B+2+=+6+
+11%2A%28+n%2B2+%29+=+n%2A%28+n%2B1+%29+%2B+46+
+11%2A6+=+4%2A5+%2B+46+
+66+=+20+%2B+46+
+66+=+66+
OK
The consecutive integers can be:
6,7, and 8
or
4,5, and 6