Question 1135926: Which 3 digit number is 11 times the sum of its digits?
Found 2 solutions by ankor@dixie-net.com, rothauserc:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Which 3 digit number is 11 times the sum of its digits?
:
100a + 10b + c = 11(a+b+c)
100a + 10b + c = 11a + 11b + 11c
100a - 11a + 10b - 11b = 11c - c
89a - b = 10c
b = 89a - 10c
looking at this we know that a has to be 1 to give a single digit value to b
b = 89 - 10c
looking at this we know that c has to be 8 to give a single digit value to b
b = 89 - 80
b = 9
therefore: 198 is the number
:
:
Check:
198 = 11(1+8+9)
198 = 11(18)
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! let the digits be xyz, then
:
1) 100x +10y +z = 11 * (x +y +z)
:
(100x +10y +z)/11 = x +y +z
:
The divisibility rule for 11 is that the alternating sum of the digits must be a multiple of 11, that is
:
x -y +z = 11k, for some integer k
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Note that 11k can be 0 or 11 only, just consider values of x, y, z
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If k = 0, then
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y = x+z
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substitute for y in equation 1 and we get
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z = 8x implies that y = 9x
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so we have 198, with x = 1 and 198 satisfies 1 -9 +8 = 0
:
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for k = 0, we have 198
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if k = 1, then
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x -y +z = 11
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y = x +z -11
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substitute for y in equation 1 and we get
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z = 8x +1, implies y = 9x -10
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if x is 1 then y is -1 and if x = 2, then z = 11
:
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there is no solution for k = 1
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