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T O P I C    R E V I E W
TchrWill Posted - 03/22/2013 : 18:15:55
What three-digit positive integer is exactly 32 times the sum of its digits?

Enjoy!
2   L A T E S T    R E P L I E S    (Newest First)
TchrWill Posted - 03/23/2013 : 18:03:19
quote:
Originally posted by Ultraglide

576



Right on Ultraglide. Might you share your solution path with us?

One of mine is as follows:

1--Lets express your 3 digit number as N = 100A + 10B + C
2--The requirement is to have 100A + 10B + C = 32(A + B + C)
3--This equality can be simplified to 31C + 22B = 68A
4--For there to be any positive integer solutions of this equation, the g.c.d. of 31 and 22 must evenly divide 68.
5--By inspection, the g.c.d.(31,22) = 1 which clearly divides 68.
6--We first seek to find a solution to 31C + 22B = 68 assuming A = 1 for now.
7--Dividing through by 22 yields 1C + 9C/22 + B = 3 + 2/22 or (9C - 2)/22 = 2 - C - B
8--(9C - 2)/22 must be an integer but we want to get the coefficient of C = 1.
9--(9C - 2)/22 x 5 = (45C - 10)/22
10--Dividing by 22 again yields 2C + 1/22 - 10/22
11--Again, (C - 10)/22 must be an integer k making C = 22k + 10
12 Substituting this back into (6) yields 31(22)k + 310 + 22B = 68 or B = -31k - 11.
13--A solution derives from k = 0 or C = 10 and B = -11.
14--Other solutions derive from C = 10 + 22t and B = -11 - 31t
15--If integer solutions exist, they must fall between (-10/22) < t < (-11/31) or
......-.4542 < t < -.3548.
16--Clearly no integers exist between these limits so A must be more than 1.
17--Letting A increase in increments of 1 and establishing the new limits for t, we eventually reach 31(50) + 22(-55) = 340 where A = 5.
18--Then, C = 50 + 22t and B = -55 - 31t.
26--The limits now become -50/22 < t < -55/31 or -2.272 < t < -1.774.
29--Clearly, t = -2 fits between these limits.
30--Therefore, C = 50 - 22(-2) = 6 and B = -55 - 31(-2) = 7.
31--Therefore, our number is 576.

Ultraglide Posted - 03/23/2013 : 16:18:22
576

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