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Question 135924: The ten's digit of a certain two digit number is twice the units digit.If the number is multiplied by the sum of its digits the result is 63.Find the number.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
Let x=the tens digit
Ane let y=the units digit
Now we are told that:
x=2y---------------------------eq1
The sum of its digits is x+y and the number is 10x+y, ok?
And we are being told that:
(x+y)(10x+y)=63 using the FOIL (First, Outer, Inner, Last) crutch we get:
10x^2+xy+10xy+y^2=63 and this equals
10x^2+11xy+y^2=63--------------------------eq2
substitute x=2y from eq1 into eq2
10(2y)^2+11(2y)(y)+y^2=63 simplify
40y^2+22y^2+y^2=63 collect like terms
63y^2=63 divide each side by 63
y^2=1 take square root of each side
y=+ or - 1 so
y=+1
and y=-1
And substituting these into eq1, we get:
x=+2, when y=+1
and
x=-2, when y=-1
So our answer is +21
CK
Lets look at +21
2 is twice 1 so first condition is met; if the number is multiplied by the sum of its digits, the result is 63---3*21 is 63 so the second condition is met
Hope this helps---ptaylor
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