How do you find two consecutive integers whose product is 783?

Question

Savannah Anderson · Accepted Answer

See a solution process below:

First, let's call the first integer: #n# Then, the next consecutive integer would be by definition: #n + 1# We can then write and solve this equation for find #n#: #n(n + 1) = 783# #n^2 + n = 783# #n^2 + n - color(red)(783) = 783 - color(red)(783)# #n^2 + n - 783 = 0# We can now use the quadratic equation to solve this problem: The quadratic formula states: For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by: #x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))# Substituting: #color(red)(1)# for #color(red)(a)# #color(blue)(1)# for #color(blue)(b)# #color(green)(783)# for #color(green)(c)# gives: #x = (-color(blue)(1) +- sqrt(color(blue)(1)^2 - (4 * color(red)(1) * color(green)(-783))))/(2 * color(red)(1))# #x = (-1 +- sqrt(1 - (-3132)))/2# #x = (-1 +- sqrt(1 + 3132))/2# #x = (-1 - sqrt(1 + 3132))/2# and #x = (-1 + sqrt(1 + 3132))/2# #x = (-1 - sqrt(3133))/2# and #x = (-1 + sqrt(3133))/2# #x ~= (-1 - 55.973)/2# and #x ~= (-1 + 55.973)/2# #x ~= (-56.973)/2# and #x ~= (54.973)/2# #x ~= (-56.973)/2# and #x ~= (54.973)/2# #x ~= -28.487# and #x ~= 27.487# As show by this answer there are not two consecutive integers which when multiplied give 783. There are however two consecutive ODD integers which when multiplied give 783: #27 * 29 = 783#

Julia Adams · Answer

Consecutive ODD numbers which give #783# are:
#27 xx 29 =783# At first glance we should see that there are no such integers.... Integers alternate between odd and even all the way along the number line. Therefore one of every two consecutive numbers will be even. The multiple of any even number is always even. #"odd" xx "even" = "even"# #783# is odd and therefore cannot be the product of an even and an odd number. However, if the question is supposed to read The product of two consecutive ODD numbers is #783#, then we can proceed as follows: If the factors of a number, such as #36# are arranged in order we would have: #color(blue)(1),2,3,color(red)(4), 6, color(red)(9),12,18,color(blue)(36)# The difference between any pairs of factors are greatest for the outer factors and smallest for the inner factors. #color(blue)(36-1 = 35)" " color(red)(9-4 - 5)" "6-6 =0# In the case of #36#, the factor exactly in the middle is #6# #sqrt36 = 6# Consecutive factors differ by #1# and will lie on either side of the square root of a number. Consider #42# Factors are #1,2,3,6,7,14,28,42# #color(white)(xxxxxxxxxxxx)uarr# #color(white)(xxxxxxxxxxx)sqrt42# #sqrt42 = 6.48# which lies between #6 and 7# In the case of #783# #sqrt783 = 27.982# This is very close to #28# which lies between consecutive odd numbers #27 and 29# Check: #27 xx 29 = 783#