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Question 146727: Hi
I am having problems figuring out 3 digit word problems. The problem is:
In a three digit number which is 31 times the sum of the digits, the units' digit is one half the sum of the other digits. If the digits are reversed, the new number obtained is 99 greater than the original number. Find the digits.
Can you help me with this?
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! let h=hundreds digit, t=tens digit, and u=units digit
"three digit number which is 31 times the sum of the digits" __ 31(h+t+u)=100h+10t+u
"the units' digit is one half the sum of the other digits" __ 2u=h+t __ 2u-h=t
"If the digits are reversed, the new number obtained is 99 greater than the original number"
__ 100u+10t+h=100h+10t+u+99
__ subtracting 10t __ 100u+h=100h+u+99 __ subtracting u+h __ 99u=99h+99 __ dividing by 99 __ u=h+1 __ u-1=h
substituting __ 2u-(u-1)=t __ u+1=t
substituting __ 31((u-1)+(u+1)+u)=100(u-1)+10(u+1)+u __ 93u=111u-90 __ 90=18u __ 5=u
substituting __ (5)-1=h __ 4=h
substituting __ (5)+1=t __ 6=t
the number is 465
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