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consecutive integers=x, x+1, x+2, x+3
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First way: Two consecutive numbers
x+(x+1)=99
2x=98
x=49
x+1=50
ANSWER: 49+50=99
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Second way: Three consecutive numbers
x+(x+1)+(x+2)=99
3x+3=99
3x=96
x=32
x+1=33
x+2=34
ANSWER:32+33+34=99
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Third way: Four consecutive numbers
x+(x+1)+(x+2)+(x+3)=99
4x+6=99
4x=93 This is not an integer, 99 can not be written with 4 consecutive integers.
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Third way:
Take an odd factor of the number (33 is a odd factor of 99)
Multiply the number by 2 and divide by the odd factor: Subtract 6 from 33 (subtract whichever is smaller from larger: The smaller of the two is also the number of terms) to get 27, add 1 to 27 and divide by 2=14
Start at 14 and write 6 (the smaller of 6 and 33) consecutive numbers:
14+15+16+17+18+19=99
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You can choose any odd factor of the number. If we had chosen 3 as factor: Subtract 3 from 66 (smaller from larger) to get 63, add 1 and divide by 2 to get 32.
Start at 32 and write 3 (the smaller of 3 and 66)consecutive numbers:
32+33+34=99 Note this is the same end result as second way above.
