Number * Sum of Remaining Four NumbersDate: 04/03/2003 at 20:33:29 From: John Subject: Adding/multiplying groups of numbers Find 5 numbers such that when each number is multiplied by the sum of the remaining 4 numbers, the following values will result: 152, 245, 297, 320, 360. Thank you! Date: 04/04/2003 at 00:11:42 From: Doctor Greenie Subject: Re: Adding/multiplying groups of numbers Hi, John - Cool problem! Lots of good mental exercise and mathematical detective work here. Let's call the 5 numbers a, b, c, d, and e. Then we have (1) a(b+c+d+e)=152 (2) b(a+c+d+e)=245 (3) c(a+b+d+e)=297 (4) d(a+b+c+e)=320 (5) e(a+b+c+d)=360 It seems that with the equations written this way, the numbers a, b, c, d, and e should be in numerical order. This can in fact be demonstrated quite easily by comparing any two consecutive equations. For example, distributing in equations (2) and (3) and subtracting equation (2) from equation (3), we have ca+cb+cd+ce = 297 ba+bc+bd+be = 245 ------------------- (c-b)(a+d+e) = 52 Since this product is positive, (c-b) must be positive, which means c is greater than b. Using other pairs of "consecutive" equations, we can similarly show that b is greater than a, d is greater than c, and e is greater than d. So we know a < b < c < d < e From equations (2) and (3), since the products are odd, we know that b and c are odd integers. Looking at those two equations from a different perspective, we see that, since the same 5 numbers occur on the left side of both equations, it is necessary to find number p, q, r, and s such that pq = 245 rs = 297 with the restriction that p+q = r+s We have the following possible factorizations of 245 and 297: 245 = 5*49 = 7*35 297 = 3*99 = 9*33 = 11*27 The only factorizations of these two numbers that satisfy the requirements are 245 = 7*35; 297 = 9*33 So we know (*) b = 7 (*) c = 9 Using these two values in either equation (2) or (3) above, we can also determine that a+d+e = 26 And from this we know that a+b+c+d+e = 42 Now let's look at equations (1), (4), and (5) again, knowing the values of b and c. We have a(16+d+e) = 152 d(16+a+e) = 320 e(16+a+d) = 360 Because a+d+e=26, we need to find a factorization of 152 into the product of two integers whose sum is 42; and likewise factorizations of 320 and 360 each into the product of two integers whose sum is 42. With some trial and error we find 152 = 4*38 320 = 10*32 360 = 12*30 And so we conclude (*) a = 4 (*) d = 10 (*) e = 12 So we have a = 4; b = 7; c = 9; d = 10; e = 12 Checking our results in the original equations: 4(7+9+10+12) = 4(38) = 152 7(4+9+10+12) = 7(35) = 245 9(4+7+10+12) = 9(33) = 297 10(4+7+9+12) = 10(32) = 320 12(4+7+9+10) = 12(30) = 360 Looking back at this solution, it appears there could have been many different paths to the answer using the same types of logical analysis.... I hope this helps. Please write back if you still have any questions about any of this. And thanks for providing me with some good mental exercise! - Doctor Greenie, 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/