Multiplying Groups of NumbersDate: 5/9/96 at 12:8:46 From: Anonymous Subject: Whole Numbers and Multiplication Please help. The following is doing me in... Arrange the nine digits 1,2,3,4,5,6,7,8 and 9 into three groups. Make one group of two digits, one group of three digits, and one group of four digits, so that when you multiply the first group of two times the second group of three you will get the third group of four. Example: 39 x 186 = 7,254 There are six other possible combinations. My 10 year old must find one. I am his mom, and to be politically correct, I am mathmatically challenged. Thanks Date: 12/11/96 at 01:02:12 From: Doctor Rob Subject: Re: Whole Numbers and Multiplication I would start by looking at the units digit of the two-digit and three-digit numbers. Let's write the multiplication problem as abc * de = fghi Then we'll consider cases, depending on the values of c and e. Neither one can be a zero, obviously. Neither one can be a 1, either, since if c = 1, then i = e, which is excluded, and if e = 1, then i = c, which is also excluded. Similarly, neither one can be a 5, because 5 times an even number ends in 0, and 5 times an odd number ends in 5, both of which are impossible. This leaves 42 possible pairs of values of c and e. Certain others can also be eliminated, since 6 times an even number ends in that same even number. This reduces the list to 36 possibilities: c e i c e i c e i c e i c e i c e i c e i c e i c e i ----- ----- ----- ----- ----- ----- ----- ----- ----- 2 3 6 2 4 8 2 7 4 2 8 6 2 9 8 3 2 6 3 4 2 3 6 8 3 7 1 3 8 4 3 9 7 4 2 8 4 3 2 4 7 8 4 8 2 4 9 6 6 3 8 6 7 2 6 9 4 7 2 4 7 3 1 7 4 8 7 6 2 7 8 6 7 9 3 8 2 6 8 3 4 8 4 2 8 7 6 8 9 2 9 2 8 9 3 7 9 4 6 9 6 4 9 7 3 9 8 2 The next step is to see which b and d can possibly go with each of these 36 possibilities. Let us try the first one, c = 2, e = 3, i = 6. Clearly b and d cannot be any of these or zero, and they must be unequal, so there are 30 possible pairs. We know that the digit h must equal the last digit of the sum of c*d, plus b*e, plus the carry from the product c*e. In this case, that is: 2*d + 3*b + 0 = 2*d + 3*b This last digit h also cannot be 0, 2, 3, 6, b, or h. We try all thirty: 1 4 1 1 5 3 1 7 7 1 8 9 1 9 1 4 1 4 4 5 2 4 7 6 4 8 8 4 9 0 5 1 7 5 4 3 5 7 9 5 8 1 5 9 3 7 1 3 7 4 9 7 5 1 7 8 7 7 9 9 8 1 6 8 4 2 8 5 4 8 7 8 8 9 2 9 1 9 9 4 5 9 5 7 9 7 1 9 8 3 Any of these with h = b or h = d or h = 0, 2, 3, or 6 is impossible. This leaves: b d h b d h b d h b d h b d h ----- ----- ----- ----- ----- 1 8 9 5 1 7 5 7 9 5 8 1 7 4 9 7 5 1 8 5 4 9 4 5 9 5 7 9 7 1 or only ten possibilities for this first case. The next step is to test the remaining values of a for each of these ten subcases. a cannot be zero, b, c, d, e, h, or i, so there are only three possibilities in each subcase. Let us try the first one, b = 1, d = 8, h = 9 (and recall c = 2, e = 3, i = 6). The three values of a are 4, 5, or 7. 412*83 = 34196, which is too big, and the other values of a will all give even bigger answers, so none of them works. This eliminates the first subcase. Next we have to try a second subcase, then a third, and so on, until we either find a solution, or else eliminate all ten of the subcases. If we eliminate all ten of the subcases, that means that the first of the 36 cases is eliminated, and we go on to the next. I know that this is a long and complicated process, but it is sure way to find all the solutions, and to prove that you have them all. I wish I could think of some useful shortcuts, but this is all that I could figure out to solve your problem. If you need more help, write back again, and we'll try to offer more. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/