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196 = 100(x d) 10(b e) 1(z c) Since our focus is on the units digit, notice that the units digit on the left side of the equation is 6 and the units digit on the right side of the equation is (z c) Thus, we know that 6 = z c Since z and c are single positive digits, let's list the possible solutions to this equation z = 9 and c = 3Feb 09, 07 · 1 at X = 0 the X terms = 0 so f(0) = d = 2 2 f(1) = 5 = abcd so abc = 7 3 f(1) = 3 = abcd ==> take out d ==> abc = 5 4 add "abc = 7" and "abc = 5" to get 2b = 12 b=6 5 f(2) = 4 = 8a 4b 2c d plug in b = 6 and d = 2 4 = 8a 24 2c 2 8a 2c = 18 ===> 4a c = 9 6 Subtract "abc = 5" from "abc = 7" to get 2a 2c = 2 a c = 1 ==> c=19 N4 26 Vol348 q X ^ ~ s Ϗǂ āH ̌ N ̂ ߂ɁA R u ` i g L m R j A X Ȃǂ̔ y H i A R s L x ȃg } g A ǎ ̎ ܂ރA { J h ێ悷 ȂǁA N ɋC t Ă l ͑ Ǝv ܂ B Ȃ̂ɁA C C A 畆 ɂ ݂⎼ ł A @ o Ėڂ Ȃǂ̃A M Ǐ A ɂɔY ܂ ꂽ A S o N o N Ă ƁA u Ȃ H v Ǝv Ƃ͂ ܂ H @ t ɑ k Ă A Ǐ ̌ Ȃ 悤 Ȃ A q X ^ ~ s Ϗǂ l 邩 ܂ B
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Jun 03, 08 · For the best answers, search on this site https//shorturlim/avXKY This is not strictly greater, as equality can occur for certain values of a,b,c,x,y,z (for example when they are all zero) Expanding terms, we get (x^2 y^2 z^2)(a^2 b^2 c^2) (x*a y*b z*c)^2 = (xb)^2 (xc)^2 (ya)^2 (yc)^2 (za)^2 (zb)^2 2xbya 2xcza 2yczb = (xb ya)^2 (xc za)^2N A r m i t a g e R d Gr e n Acres Rd A m a z o n P a r k w a y L o ra n e H w y L o r a n e O H w y T i m b e r l i n e D r Warren S t G i mp le H i l d H dsn al e o L n Ed C o n e B lv d W el c o m e W a R os e vlt Bl d F i r e Hio d a W a g n e r S Cambon A ve T r e v o n Maxwell Rd S t B r a e B u r Dr D e l w o o d D r
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