4(ABCD) = DCBA
Date: 06/15/97 at 17:36:44 From: Ziva Subject: Math problem Find a four digit number such that: ABCD * 4 ------ DCBA
Date: 06/17/97 at 16:27:04 From: Doctor Rob Subject: Re: Math problem This is one of those digital logic problems. You must use the facts that the digits are integers from 0 to 9 and what you know about arithmetic to figure this one out. These problems can be tricky and/or tedious, but this one isn't too bad. Start with some inequalities: 1 <= A <= 9 (since ABCD is a four-digit number, A can't be zero) 0 <= B <= 9 0 <= C <= 9 4 <= D <= 9 (D must be at least 4*A which is at least 4) Then we have some equations: 4*D = 10*z + A (z is the carry to the tens column) 4*C + z = 10*y + B (y is the carry to the hundreds column) 4*B + y = 10*x + C (x is the carry to the thousands column) 4*A + x = D There are more inequalities: 0 <= z <= 3 0 <= y <= 3 0 <= x <= 3 (because the multiplier is 4, the largest a carry can be is 3) Now we combine the first and last equations by multiplying the last by 4 and substituting for 4*D in the first equation: 4*(4*A + x) = 10*z + A or 4*x = 10*z - 15*A = 5*(2*z - 3*A) Since 5 divides the number on the right, 5 must divide 4*x, and so 5 divides x. But x is small, from 0 to 3, so x = 0 is forced. Thus 2*z - 3*A = 0, so 2*z = 3*A. Similarly, 3 divides z, so put z = 3*w, so A = 2*w. Then D = 8*w, and 4 <= D <= 9. Clearly D = 8, so w = 1, A = 2, and z = 3. Similarly, using the middle two equations, we can determine B, C, and y. This gives you the answer you seek. You can finish the rest. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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