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The sum of a three-digit number and a one-digit number is 160. The product of the two number is 624. What are the two numbers?

Respuesta :

NobleAlexander
Let's look at all the potential numbers: 

3-Digit Numbers: 151, 152, 153, 154, 155, 156, 157, 158, 159
1-Digit Numbers: 9, 8, 7, 6, 5, 4, 3, 2, 1

See what numbers the product is divisible by via either a calculator or long division. (Divisible means that when you divide it by said number it comes out as a whole number. Example 624/1 = 624) Once you've done that for the 1 digit numbers, also do the same process for the 3 digit numbers. 

Divisible 1-Digit Numbers: 8, 6, 4, 3, 2, 1
Divisible 3-Digit Numbers: 156

The 3-Digit Number has to be 156. So the complimentary number with it is 4. (Because 160-156 = 4) 

So our answers are: 
156, and 4


apologiabiology
here is the legit way to solve
the numbers are x and y
x+y=160
product is 624
xy=624

x+y=160
minus x from both sides
y=160-x
sub that for y
x(160-x)=624
distribute
160x-x^2=624
multiply both sides by -1
x^2-160x=-624
add 624 both sides
x^2-160x+624=0

use quadratic formula
for
ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]

a=1
b=-160
c=624

x=[tex] \frac{-(-160)+/- \sqrt{(-160)^2-4(1)(624)} }{2(1)} [/tex]
x=[tex] \frac{160+/- \sqrt{25600-2496} }{2} [/tex]
x=[tex] \frac{160+/- 152 }{2} [/tex]
x=(312)/2 or 8/2
x=156 or 4
those are the 2 numbers


the way that doesn't require as much work is
there are only 9 possible one digit numbers
it cannot have decimpals when divided
trial and error
624/1=624, 624+1=625, not 160
624/2=312, 312+2=314, not 160
624/3=208, 208+3=211, not 160
624/4=156, 156+4=160, great


anyway, the numbers are 4 and 156

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