SOLUTION: the sum of the digits of a number of three digits is 18. if the number is divided by the sum of the ten's and hundred's digits. the quotient is 38 and the remainder is 1. if 99 is
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Question 819681: the sum of the digits of a number of three digits is 18. if the number is divided by the sum of the ten's and hundred's digits. the quotient is 38 and the remainder is 1. if 99 is added to the number, the digits will be revised. find the number. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Let a = the 100's digit
Let b = the 10's
Let c = the units
then
100a+10b+c = "the number"
:
the sum of the digits of a number of three digits is 18.
a + b + c = 18
:
if the number is divided by the sum of the ten's and hundred's digits.
the quotient is 38 and the remainder is 1. = 38
Multiply by (a+b)
100a + 10b + c - 1 = 38(a+b)
100a + 10b + c - 1 = 38a + 38b
100a - 38a + 10b - 38b + c = 1
62a - 28b + c = 1
:
if 99 is added to the number, the digits will be revised.
100a+10b+c+99 = 100c+10b+a
100a - a + 10b - 10b + c - 100c = -99
99a - 99c = -99
Divide by 99
a - c = -1
or
a + 1 = c
In the 1st equation, replace c with (a+1)
a + b + (a+1) = 18
2a + b = 17
In the 2nd equation, replace c with (a+1)
62a - 28b + a + 1 = 1
63a - 28b = 1 - 1
63a - 28b = 0
Use elimination here multiply by 2a + b = 17 by 28, add to the above
56a + 28b = 476
63a - 28b = 0
---------------adding eliminates b find a
119a = 476
a = 476/119
a = 4
then
c = 4 + 1
c = 5
Find b
4 + b + 5 = 18
b = 18 - 9
b = 9
:
495 is the number
:
See if the works out when you add 99, does it reverse the digits?
495 + 99 = 594
