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Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, the smallest integer is:

Answer
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Hint: Proper conversion of wordings to equation form is must. It is easy to solve such problems with variables, as they can be interrelated easily like b=a+1 to denote they are consecutive in nature.

Complete step-by-step answer:
Here, three consecutive positive integers are such that the sum of the first and product of the other two is 46.
Let us consider these integers as a,b and c such that
b=a+1c=b+1
Or, c=(a+1)+1=a+2... (1)
Now, as per the given conditions, we have
a2+bc=46, where a is first integer, b is second integer and c is third integer.
Substituting the values of b and c from equation (1), we get
a2+bc=46a2+(a+1)(a+2)=46a2+(a(a+2)+1(a+2))=46a2+(a2+2a+a+2)=46a2+a2+3a+2=462a2+3a+2=46
By transposing values, we get
2a2+3a+2=462a2+3a+246=02a2+3a44=0
Applying middle-term split method, we get
2a2+3a44=02a2+11a8a44=0a(2a+11)4(2a+11)=0(a4)(2a+11)=0
Thus, solving the above equation, we get
(a4)(2a+11)=0, i.e.,
(a4)=0a=4
And,
(2a+11)=02a=11
On cross-multiplication, we get
a=112
But as per the given conditions, a is a positive integer, so a=112 is not applicable.
Thus, a=4 and from equation (1), we can calculate other two integers, i.e.,
b=a+1b=4+1b=5
Similarly,
c=a+2c=4+2c=6
Hence, the smallest integer out of three consecutive integers, we have a=4.

Note: This type of problem can also be solved with the hit and trial method. Considering values of those three consecutive integers one by one, trying to fit it in given conditions.
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