How do you find two consecutive positive odd integers whose product is 143?

3 Answers
Feb 6, 2016

Answer:

#143 = 11 xx 13#

Explanation:

Notice that consecutive odd integers will differ by #2#, so we could call the even number between them #n#, making the two integers #n-1# and #n+1#.

So:

#143 = (n-1)(n+1) = n^2-1#

Add #1# to both ends and transpose to get:

#n^2 = 144 = 12^2#

So #n = +-12#

Since we are told that the integers are positive, use #n=12# to find the pair of integers: #11, 13#.

Alternatively, just find factors of #143#, trying prime numbers in turn: #2# (no), #3# (no), #5# (no), #7# (no), #11# (yes). Then #143/11 = 13# - we have found our pair of numbers.

Feb 6, 2016

Answer:

11 and 13

Explanation:

Proceed as below.
Let #(2n+1)# be the first odd integer, where #n# is any positive integer. Second number being consecutive is #(2n+3)#
Rest of the steps are same as followed by others.

It is by convention, as #2n+1# is always odd.

Feb 6, 2016

Answer:

Expanding on AO8's answer: This approach guarantees that any number we look at is odd:#" "#Solution: 11 & 13

Explanation:

We need to be able to guarantee that we are dealing with odd numbers.

Let #n# be any number where #n>0#

Suppose n is even. Then #2n# is also even.
Suppose n is odd. Then #2n# is also even.

But in both cases #2n+1# will be odd.

Let the first odd number be #2n+1#
Then the second odd number will be #2n+3#

We are given that the product is 143 so write the equation as:

#(2n+1)(2n+3)=143#

This is saying exactly the same thing as the others but in a slightly different way! So we now have:

#4n^2+8n-140=0#

Divide by 4

#n^2+2n-35" "=" "(n+7)(n-5)" "=" "0#

So for my conditions #n=-7" or "+5#

#n=-7" does not work so we use " n=5#

Thus the first number is: #(2n+1)" "->" "2(5)+1=11#

So the second number is #11+2=13#