Solution:
True.
Justification:
At least one out of every three consecutive positive integers is divisible by 2.
Therefore, The product of three consecutive positive integers is divisible by 2.
At least one out of every...
Solution:
Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.
By Euclid’s division lemma, we have
a = bq + r; 0 ≤...
Let the three consecutive integers bex, x+1and x+2.
According to the question, x+x+1+x+2 = 51
=> 3x+3=51
=> 3x+3-3=51-3 [Subtracting 3 from both sides]
=> 3x=48
=> 3x/3 = 48/3 [Dividing both sides by 3]
X=16
Hence, first...
Let the three consecutive integers be x, x+1and x+2.
According to the question, 2x+3(x+1)+4(x+2)=74
=> 2x+3x+3+4x+8=74
=> 9x+11=74
=>9x+11-11=74-11 [Subtracting 11fromboth sides]
=> 9x=63
=> 9x/9=63/9 [Dividing both sides by9]
=> x=7
Hence first integer = 7, second integer=7...
True, because n(n+1)(n+2) will always be divisible by 6, as least one of the factors will be divisible by 2 and at one of the factors will be divisible by...
Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x + 2.
Since both the integers are smaller than 10,
x + 2 <...
Let x be the smaller of the two consecutive even positive integers. Then, the other integer is x + 2.
Since both the integers are larger than 5,
x > 5 ..........................
