Numbers Identical in Base 10 and Base 3Date: 02/14/2001 at 07:51:49 From: Gwendolyn Scott Subject: Base ten word problem Please help me with this word problem! Find all base-10 numbers that, when multiplied by eight, have exactly the same digits as their base-3 counterparts. I really need an explanation of how you got the answer. I don't know where to start. Thanks. Date: 02/14/2001 at 11:25:32 From: Doctor Rob Subject: Re: Base ten word problem Thanks for writing to Ask Dr. Math, Gwendolyn. I think we can assume that N >= 0. If N were negative, then -N would have the same property. Let N be the number in question. If N has d digits, then 8*N < 8*10^d On the other hand, the smallest base-3 number corresponding to N is 10^(d-1) (base-10) = 101^(d-1) (base-3) > 100^(d-1) (base-3) = 10^(2*d-2) (base-3) Now 8*10^d < 10^(2*d-2) iff 8 < 10^(d-2) iff log(8) < d - 2 iff 0.9031 < d - 2 iff 2.9031 < d That means when d >= 3, you get N < 8*10^d < 10^(2*d-2) < N a contradiction, so no such N can exist. That means we can restrict attention to N < 100. For 1-digit numbers, you would have 8*N = 10*a + b (0 <= a <= 2, 0 <= b <= 2), N = 3*a + b 7*N = 7*a N = a a = 3*a + b -2*a = b This implies that a = b = 0, so N = 0. That is one solution. For 2-digit numbers, you would have 8*N = 100*a + 10*b + c N = 9*a + 3*b + c 7*N = 91*a + 7*b N = 13*a + b Here a, b, and c are all in the set {0,1,2}. The above imply that 0 = -4*a + 2*b + c c = 4*a - 2*b Try all nine possibilities for a and b. Throw out those for which c isn't 0, 1, or 2. Those will give you the other positive solutions. Take their negatives to get all the negative solutions, and you will have a complete list. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/